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;
long long n, m, a[100010], l[100010], r[100010], LS[100010], T;
int G[100010][10];
struct MAT {
  int data[10][10];
  MAT() { memset(data, 0, sizeof(data)); }
  MAT operator*(const MAT& ma) const {
    MAT ret;
    for (int i = 1; i <= 3; ++i)
      for (int j = 1; j <= 3; ++j)
        for (int k = 1; k <= 3; ++k)
          ret.data[i][j] =
              (ret.data[i][j] + 1ll * data[i][k] * ma.data[k][j]) % 1000000007;
    return ret;
  }
};
MAT A;
int main() {
  scanf("%lld%lld", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%lld%lld", &a[i], &l[i], &r[i]);
    LS[++T] = l[i], LS[++T] = r[i] + 1;
  }
  LS[++T] = 2, LS[++T] = m + 1;
  sort(LS + 1, LS + T + 1);
  T = unique(LS + 1, LS + T + 1) - LS - 1;
  for (int i = 1; i <= n; ++i) {
    l[i] = lower_bound(LS + 1, LS + T + 1, l[i]) - LS;
    r[i] = lower_bound(LS + 1, LS + T + 1, r[i] + 1) - LS;
    G[l[i]][a[i]]++, G[r[i]][a[i]]--;
  }
  for (int i = 1; i <= 3; ++i) A.data[i][i] = 1;
  for (int i = 1; i < T; ++i) {
    for (int j = 1; j <= 3; ++j) G[0][j] += G[i][j];
    MAT I;
    for (int j = 1; j <= 3; ++j)
      for (int k = max(1, j - 1); k <= min(3, j + 1); ++k)
        if (!G[0][k]) I.data[j][k] = 1;
    long long len = LS[i + 1] - LS[i];
    for (; len; len >>= 1, I = I * I)
      if (len & 1) A = A * I;
  }
  printf("%d", A.data[2][2]);
  return 0;
}
