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#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <list>
#include <ctime>
#include <sstream>
#include <queue>
#include <stack>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
using namespace std;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef long long ll;
typedef short int sint;
#define FOR(x, b, e) for(int x=(b); x<=(e); ++x)
#define FORD(x, b, e) for(int x=((int)(b))-1; x>=(e); --x)
#define REP(x, n) for(int x=0; x<(n); ++x)
#define ALL(c) c.begin(),c.end()
#define SIZE(x) ((int)((x).size()))
#define PB push_back
#define ST first
#define ND second
#define mp(x,y) make_pair(x,y)
#define DEBUG 1
#define debug(x) {if (DEBUG)cerr <<#x <<" = " <<x <<endl; }
#define debugv(x) {if (DEBUG) {cerr <<#x <<" = "; FOREACH(it, (x)) cerr <<*it <<", "; cout <<endl; }}
#define REMAX(a,b) (a)=max((a),(b));
#define REMIN(a,b) (a)=min((a),(b));
#define wez(n) int (n); scanf("%d",&(n));
#define wez2(n,m) int (n),(m); scanf("%d %d",&(n),&(m));

bool dbg = true;
const int N = 101;
int p[N], q[N], c[N];
int all[2 * N], cnt;
int n, m;

bool ok(int pocz, int kon) {
    int sumMust = 0;
    REP(i, n) {
        if (q[i] < pocz || p[i] > kon) continue;
        int poza = 0;
        if (p[i] >= pocz && q[i] <= kon) {
            poza = 0;
        } else if (p[i] <= pocz && kon <= q[i]) {
            poza = pocz - p[i] + q[i] - kon;
        } else if (p[i] < pocz) {
            poza = pocz - p[i];
        } else { // q[i] > kon
            poza = q[i] - kon;
        }
        if (poza < c[i]) {
            sumMust += (c[i] - poza);
        }
    }
    int moge = (kon - pocz + 1) * m;
    if (sumMust > moge) return false;
    return true;
}

int main() {
    scanf("%d %d", &n, &m);
    REP(i, n) {
        scanf("%d %d %d", &p[i], &q[i], &c[i]);
        --q[i];
        all[cnt++] = p[i];
        all[cnt++] = q[i];
    }
    sort(all, all + cnt);
    cnt = unique(all, all + cnt) - all;
    REP(i, cnt) {
        FOR(j, i, cnt - 1) {
            if (!ok(all[i], all[j])) {
                printf("NIE\n");
                return 0;
            }
        }
    }
    printf("TAK\n");
    return 0;
}