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#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef long double LD;
typedef pair < int, int > PII;
typedef pair < LL, LL > PLL;
typedef pair < LD, LD > PDD;

#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define dbl(k, x) fixed << setprecision(k) << (x)

template < typename _T > inline void _DBG(const char *s, _T x) { cerr << s << " = " << x << "\n"; }
template < typename _T, typename... args > void _DBG(const char *s, _T x, args... a) { while(*s != ',') cerr << *s++; cerr << " = " << x << ','; _DBG(s + 1, a...); }

#ifdef LOCAL
#define _upgrade ios_base::sync_with_stdio(0);
#define DBG(...) _DBG(#__VA_ARGS__, __VA_ARGS__)
#else
#define _upgrade ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define DBG(...) (__VA_ARGS__)
#define cerr if(0) cout
#endif

// ********************** CODE ********************** //

const int N = 1e4 + 7;
const int M = 1e9 + 7;

LL fast(LL a, LL n)
{
    LL w = 1;
    while(n)
    {
        if(n & 1) w = w * a % M;
        a = a * a % M;
        n /= 2;
    }
    return w;
}

LL fac[N], rev[N], pre[N];

LL choose(int n, int k)
{
    assert(0 <= k && k <= n);
    return fac[n] * rev[k] % M * rev[n - k] % M;
}

int n, r[3], x[3][N];

LL solve1(int t1)
{
    LL ans = 0;
    for(int i = 0; i <= r[t1]; i++)
        ans = (ans + choose(n, i)) % M;
    return ans;
}

LL solve2(int t1, int t2)
{
    //DBG(t1, t2);
    LL ans = 0;
    
    int l[2] = {0, 0};
    for(int i = 0; i < n; i++)
        l[x[t1][i] != x[t2][i]]++;

    pre[0] = 0;
    for(int i = 1; i <= l[1] + 1; i++)
        pre[i] = (pre[i - 1] + choose(l[1], i - 1)) % M;
    
    for(int i = 0; i <= min({r[t1], r[t2], l[0]}); i++)
    {
        LL tmp = 0;
        // j + i <= r[t1] && l[1] - j + i <= r[t2]
        // j <= r[t1] - i && l[1] - r[t2] + i <= j
        int a = max(0, l[1] - r[t2] + i), b = min(r[t1] - i, l[1]);
        if(a <= b)
            tmp = (tmp + pre[b + 1] - pre[a] + M) % M;
        ans = (ans + tmp * choose(l[0], i)) % M;
        //DBG(tmp, choose(l[0],i), ans);
    }

    return ans;
}

int sum[N][N];
LL tmp0[N], tmp1[N], tmp2[N];

LL solve3()
{
    LL ans = 0;

    int l[4] = {0, 0, 0, 0};
    for(int i = 0; i < n; i++)
    {
        if(x[0][i] == x[1][i] && x[1][i] == x[2][i])
            l[3]++;
        else if(x[1][i] == x[2][i])
            l[0]++;
        else if(x[0][i] == x[2][i])
            l[1]++;
        else
            l[2]++;
    }

    for(int i = 0; i <= l[0]; i++)
        tmp0[i] = choose(l[0], i);
    for(int i = 0; i <= l[1]; i++)
        tmp1[i] = choose(l[1], i);
    for(int i = 0; i <= l[2]; i++)
        tmp2[i] = choose(l[2], i);
    for(int i = 0; i <= l[1]; i++)
    {
        for(int j = 0; j <= l[2]; j++)
        {
            sum[i + j + 1][i - j + l[2] + 1] = tmp1[i] * tmp2[j] % M;
        }
    }
    for(int i = 1; i <= l[1] + l[2] + 1; i++)
    {
        for(int j = 1; j <= l[1] + l[2] + 1; j++)
        {
            LL y = (LL)sum[i][j] + sum[i - 1][j] + sum[i][j - 1] + M - sum[i - 1][j - 1];
            sum[i][j] = y % M;
        }
    }

    //DBG(l[0], l[1], l[2], l[3]);

    for(int i = 0; i <= min({r[0], r[1], r[2], l[3]}); i++)
    {
        LL tmp = 0;
        int c0 = r[0] - l[0] - i;
        int c1 = r[1] - l[1] - i;
        int c2 = r[2] - l[2] - i;

        for(int x0 = 0; x0 <= l[0]; x0++)
        {
            int a = min(l[1] + l[2], c0 + x0);
            int b1 = max(0, x0 - c1 + l[2]), b2 = min(l[1] + l[2], c2 - x0 + l[2]);
            if(b1 <= b2 && a >= 0)
                tmp = (tmp + tmp0[x0] * (sum[a + 1][b2 + 1] - sum[a + 1][b1])) % M;
            //DBG(i, a, b1, b2, x0, tmp);
        }

        ans = (ans + tmp * choose(l[3], i)) % M;
    }
    return ans;
}

int main()
{
    _upgrade
    
    cin >> n;
    for(int i = 0; i < 3; i++)
    {
        cin >> r[i];
        string s; cin >> s;
        for(int j = 0; j < n; j++)
            x[i][j] = s[j] - '0';
    }

    fac[0] = 1;
    for(int i = 1; i <= n; i++)
        fac[i] = (LL)i * fac[i - 1] % M;
    
    rev[n] = fast(fac[n], M - 2);
    for(int i = n; i; i--)
        rev[i - 1] = (LL)i * rev[i] % M;

    LL ans = solve1(0) + solve1(1) + solve1(2);
    //DBG(ans);
    ans -= solve2(0, 1) + solve2(0, 2) + solve2(1, 2);
    //DBG(ans);
    ans += solve3();
    ans = (ans % M + M) % M;
    //DBG(ans);
    cout << ans << "\n";

	return 0;
}