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 int modo = 1000000007;
struct matrix {
  int a[3][3];
  matrix(int x = 1) {
    a[0][0] = x;
    a[0][1] = 0;
    a[0][2] = 0;
    a[1][0] = 0;
    a[1][1] = x;
    a[1][2] = 0;
    a[2][0] = 0;
    a[2][1] = 0;
    a[2][2] = x;
  }
  friend matrix operator*(const matrix &x, const matrix &y) {
    int i, j, k;
    matrix c(0);
    for (i = 0; i < 3; i++) {
      for (j = 0; j < 3; j++) {
        for (k = 0; k < 3; k++) {
          c.a[i][k] = (c.a[i][k] + (long long)x.a[i][j] * y.a[j][k]) % modo;
        }
      }
    }
    return c;
  }
};
matrix now;
long long c[20005];
long long l[10005], r[10005];
int d[10005];
int val[5][20005];
matrix power(matrix x, long long y) {
  if (y == 0) return matrix();
  matrix t = power(x, y / 2);
  t = t * t;
  if (y % 2 == 1) t = t * x;
  return t;
}
int main() {
  now.a[0][1] = 1;
  now.a[1][0] = 1;
  now.a[1][2] = 1;
  now.a[2][1] = 1;
  int n;
  long long m;
  cin >> n >> m;
  int i;
  int cnt = 0;
  for (i = 0; i < n; i++) {
    cin >> d[i] >> l[i] >> r[i];
    d[i]--;
    l[i]--;
    c[cnt++] = l[i];
    c[cnt++] = r[i];
  }
  c[cnt++] = 1;
  c[cnt++] = m;
  sort(c, c + cnt);
  cnt = unique(c, c + cnt) - c;
  for (i = 0; i < n; i++) {
    l[i] = lower_bound(c, c + cnt, l[i]) - c;
    r[i] = lower_bound(c, c + cnt, r[i]) - c;
  }
  for (i = 0; i < n; i++) {
    val[d[i]][l[i]]++;
    val[d[i]][r[i]]--;
  }
  for (i = 1; i < cnt; i++) {
    int j;
    for (j = 0; j < 3; j++) {
      val[j][i] += val[j][i - 1];
    }
  }
  matrix s;
  for (i = 1; i < cnt; i++) {
    matrix t = now;
    int j;
    for (j = 0; j < 3; j++) {
      if (val[j][i - 1] > 0) {
        int k;
        for (k = 0; k < 3; k++) {
          t.a[k][j] = 0;
        }
      }
    }
    s = s * power(t, c[i] - c[i - 1]);
  }
  printf("%d\n", s.a[1][1]);
  return 0;
}
