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#include <cstdio>
#include <queue>
using namespace std;

int child[500001][2]; // left and right children of x
int intervals[2000001]; // >1'000'000 are single nodes
int parents[500001];
int totaldrips[500001];
int factorials[500001];
int visited[500001];

const int MAX_IDX = 1048575; // 2^20 - 1
const long long MODULO = 1'000'000'007LL;

void all_factorials(int n) {
    long long res = 1;
    factorials[0] = 1;
    factorials[1] = 1;
    for (int i = 2; i <= n; ++i) {
        res *= i;
        res %= 1'000'000'007LL;
        factorials[i] = int(res);
    }
}

long long odw_modulo(long long k) {
    long long r = MODULO;
    long long newr = k;
    long long t = 0;
    long long newt = 1;
    long long temp;

    while (newr != 0) {
        long long quotient = r / newr;
        temp = newt;
        newt = t - (quotient * newt);
        t = temp;

        temp = newr;
        newr = r - (quotient * newr);
        r = temp;
    }

    while (t < 0) {
        t += MODULO;
    }
    return t;
}

void setInterval(int l, int r, int cl, int cr, int x, int n) {
//    printf("SETTING INTERVAL %8d %8d    | %8d %8d  | to %d\n", cl, cr, l, r, n);
    // l, r - full required interval
    // cl, cr - interval considered in this iteration
    // x - tree index
    // n - value to set
    if (cl >= l && cr <= r) {
//        printf("   FULL INTERVAL COVERED\n");
        intervals[x] = n;
    } else if (cl > r || cr < l) {
//        printf("   NO COMMON POINTS\n");
        // no common points
        return;
    } else {
//        printf("   SPLITTING\n");
        // must set right and left
        int lhalf_end = (cl + cr) / 2;
        int lhalf_idx = x * 2 + 1;
        if (intervals[x] != -2) {
            intervals[lhalf_idx] = intervals[x];
            intervals[lhalf_idx+1] = intervals[x];
            intervals[x] = -2;
        }
        setInterval(l, r, cl, lhalf_end, lhalf_idx, n);
        setInterval(l, r, lhalf_end+1, cr, lhalf_idx+1, n);
    }
}

void setIntervalStart(int l, int r, int n) {
    setInterval(l, r, 0, MAX_IDX, 0, n);
}

int getValue(int l, int r, int x, int k) {
//    printf("CHECKING INTERVAL %8d %8d   (val: %d) | %d | \n", l, r, intervals[x], k);
    // l, r - current searched interval
    // x - tree index
    // k - required key
    if (intervals[x] != -2) {
//        printf("WHOLE INTERVAL HAS VALUE %d\n", intervals[x]);
        return intervals[x];
    }
    // else go deeper
    int lhalf_end = (l + r) / 2;
    int lhalf_idx = x * 2 + 1;
    if (k <= lhalf_end) {
        return getValue(l, lhalf_end, lhalf_idx, k);
    } else {
        return getValue(lhalf_end+1, r, lhalf_idx+1, k);
    }
}

int getValueStart(int k) {
    return getValue(0, MAX_IDX, 0, k);
}

int main() {
    int n;
    scanf("%d", &n);
    all_factorials(n);
    intervals[0] = -1; // -1 = floor, -2 = split
    for (int i = 0; i < n; ++i) {
        int l, r;
        scanf("%d %d", &l, &r);
        child[i][0] = getValueStart(l-1);
        child[i][1] = getValueStart(r-1);
        if (child[i][0] == child[i][1]) {
            child[i][1] = -1;
        }
//        parents[child[i][0]] ++;
//        parents[child[i][1]] ++;
        setIntervalStart(l-1, r-1, i);

//        printf("PLATFORM %d: L: %d, R: %d\n", i, child[i][0], child[i][1]);
    }


    // TODO: FIGURE OUT HOW TO DO LINEAR TOTALDRIP COMPUTATION
    queue<int> bfs;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < i; ++j) {
            visited[j] = 0;
        }

        int mc1 = child[i][0];
        int mc2 = child[i][1];
        if (mc1 != -1 && !visited[mc1]) {
            visited[mc1] = true;
            bfs.push(mc1);
        }
        if (mc2 != -1 && !visited[mc2]) {
            visited[mc2] = true;
            bfs.push(mc2);
        }

        while (!bfs.empty()) {
            int c = bfs.front();
            bfs.pop();
            if (c == -1) {
                continue;
            }
//            printf("%d gets a drip from %d\n", c, i);
            totaldrips[c]++;
            int c1 = child[c][0];
            int c2 = child[c][1];
            if (c1 != -1 && !visited[c1]) {
                visited[c1] = true;
                bfs.push(c1);
            }
            if (c2 != -1 && !visited[c2]) {
                visited[c2] = true;
                bfs.push(c2);
            }
        }
    }

    /* TODO: FIGURE OUT HOW TO DO LINEAR TOTALDRIP COMPUTATION
    queue<int> starting_points;

    for(int i = 0; i < n; ++i) {
        if(parents[i] == 0) {
            starting_points.push(i);
        }
    }

    while (!starting_points.empty()) {
        int t = starting_points.front();
        starting_points.pop();
        int c = child[t][0];
        if (c != -1) {
            totaldrips[c] = 1 + totaldrips[t];
        }
    }
     */

//    for (int i = 0; i < n; ++i) {
//        printf("TOTALDRIPS %d: %d\n", i, totaldrips[i]);
//    }


    long long totalresult = 0;
    // each kranik will be turned on 1/((1+totaldrips)!) times on average
    // so multiplying by all permutations
    // n! / (k - n + 1)!
    for (int i = 0; i < n; ++i) {
        long long revq = 1+totaldrips[i]; // actual contribution is 1/revq
        long long q = odw_modulo(revq);
        totalresult += q;
        totalresult %= MODULO;
        /*
        int conq = factorials[1+totaldrips[i]];
        int contributes = factorials[totaldrips[i]];
        printf("%d will contribute %d out of %d times\n", i, contributes, conq);
        */
    }

    printf("%lld\n", totalresult);
    return 0;
}