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;
struct Matrix {
  long long c[3][3];
  void clear() { memset(c, 0, sizeof(c)); }
  void zero() {
    c[0][0] = c[1][1] = c[2][2] = c[0][1] = c[1][0] = c[1][2] = c[2][1] = 1;
  }
  Matrix operator*(const Matrix &p) const {
    Matrix ans;
    ans.clear();
    for (int i = 0; i < 3; i++)
      for (int k = 0; k < 3; k++)
        for (int j = 0; j < 3; j++)
          ans.c[i][j] =
              (ans.c[i][j] + c[i][k] * p.c[k][j] % 1000000007) % 1000000007;
    return ans;
  }
};
struct node {
  int a;
  long long l, r;
} a[10005];
long long b[20010], c[3][20010], f[3];
Matrix qkpow(Matrix x, long long y) {
  Matrix v;
  v.clear();
  v.c[0][0] = v.c[1][1] = v.c[2][2] = 1;
  while (y) {
    if (y & 1) v = v * x;
    x = x * x, y >>= 1;
  }
  return v;
}
int n, num;
long long m;
int main() {
  scanf("%d %lld", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d %lld %lld", &a[i].a, &a[i].l, &a[i].r);
    a[i].a--;
    b[++num] = a[i].l - 1;
    b[++num] = a[i].r;
  }
  b[++num] = 1, b[++num] = m;
  sort(b + 1, b + num + 1);
  num = unique(b + 1, b + num + 1) - b - 1;
  for (int i = 1; i <= n; i++) {
    int l = lower_bound(b + 1, b + num + 1, a[i].l) - b;
    int r = lower_bound(b + 1, b + num + 1, a[i].r) - b;
    c[a[i].a][l]++, c[a[i].a][r + 1]--;
  }
  Matrix ans;
  ans.clear();
  ans.c[1][0] = 1;
  for (int i = 2; i <= num; i++) {
    Matrix v;
    v.clear(), v.zero();
    for (int j = 0; j < 3; j++) {
      f[j] += c[j][i];
      if (f[j]) v.c[j][0] = v.c[j][1] = v.c[j][2] = 0;
    }
    v = qkpow(v, b[i] - b[i - 1]);
    ans = v * ans;
  }
  printf("%lld", ans.c[1][0]);
  return 0;
}
