You are running through a rectangular field. This field can be represented as a matrix with 3 rows and m columns. (i, j) denotes a cell belonging to i-th row and j-th column.

You start in (2, 1) and have to end your path in (2, m). From the cell (i, j) you may advance to:

  * (i - 1, j + 1) — only if i > 1, 
  * (i, j + 1), or 
  * (i + 1, j + 1) — only if i < 3. 



However, there are n obstacles blocking your path. k-th obstacle is denoted by three integers ak, lk and rk, and it forbids entering any cell (ak, j) such that lk ≤ j ≤ rk.

You have to calculate the number of different paths from (2, 1) to (2, m), and print it modulo 109 + 7.

Input

The first line contains two integers n and m (1 ≤ n ≤ 104, 3 ≤ m ≤ 1018) — the number of obstacles and the number of columns in the matrix, respectively.

Then n lines follow, each containing three integers ak, lk and rk (1 ≤ ak ≤ 3, 2 ≤ lk ≤ rk ≤ m - 1) denoting an obstacle blocking every cell (ak, j) such that lk ≤ j ≤ rk. Some cells may be blocked by multiple obstacles.

Output

Print the number of different paths from (2, 1) to (2, m), taken modulo 109 + 7. If it is impossible to get from (2, 1) to (2, m), then the number of paths is 0.

Example

Input

2 5
1 3 4
2 2 3


Output

2
#include <bits/stdc++.h>
using namespace std;
const long long kMod = 1e9 + 7;
long long n, m, ways[3][3], trans[3][3], cntBlocked[3];
vector<tuple<long long, long long, long long> > ev;
void matrixProd(long long a[3][3], long long b[3][3]) {
  long long c[3][3];
  for (long long i = 0; i < 3; i++) {
    for (long long j = 0; j < 3; j++) {
      c[i][j] = 0;
      for (long long k = 0; k < 3; k++) {
        c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % kMod;
      }
    }
  }
  for (long long i = 0; i < 3; i++) {
    for (long long j = 0; j < 3; j++) {
      a[i][j] = c[i][j];
    }
  }
}
void matrixPow(long long pw) {
  while (pw) {
    if (pw % 2 == 1) {
      matrixProd(ways, trans);
    }
    matrixProd(trans, trans);
    pw /= 2;
  }
}
void buildTrans() {
  for (long long i = 0; i < 3; i++) {
    for (long long j = 0; j < 3; j++) {
      if (abs(i - j) <= 1 && cntBlocked[j] == 0) {
        trans[i][j] = 1;
      } else {
        trans[i][j] = 0;
      }
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m;
  for (long long i = 1, a, l, r; i <= n; i++) {
    cin >> a >> l >> r;
    ev.emplace_back(l - 1, 1, a);
    ev.emplace_back(r, -1, a);
  }
  sort(ev.begin(), ev.end());
  buildTrans();
  ways[0][1] = 1;
  matrixPow(get<0>(ev[0]) - 1);
  for (long long i = 0; i < (long long)ev.size(); i++) {
    long long crtIdx, typeEv, idxLine;
    tie(crtIdx, typeEv, idxLine) = ev[i];
    long long nxtIdx = (i + 1 < (long long)ev.size() ? get<0>(ev[i + 1]) : m);
    cntBlocked[idxLine - 1] += typeEv;
    buildTrans();
    matrixPow(nxtIdx - crtIdx);
  }
  cout << ways[0][1] << "\n";
}
