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 N = 1e4 + 7, mod = 1e9 + 7;
struct matrix {
  int a[3][3];
  inline void Clear() { memset(a, 0, sizeof(a)); }
  inline void init() {
    Clear();
    a[0][0] = a[1][1] = a[2][2] = 1;
  }
  inline void zero() {
    init();
    a[0][1] = a[1][0] = a[1][2] = a[2][1] = 1;
  }
};
struct node {
  long long l, r;
  int a;
} a[N];
int n, num;
long long m, b[N << 1], c[3][N << 1], sc[3];
matrix operator*(matrix a, matrix b) {
  matrix res;
  res.Clear();
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
      for (int k = 0; k < 3; k++)
        res.a[i][j] = (res.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;
  return res;
}
matrix fpow(matrix a, long long b) {
  matrix res;
  res.init();
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
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,
        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.a[1][0] = 1;
  for (int i = 2; i <= num; i++) {
    matrix t;
    t.zero();
    for (int j = 0; j < 3; j++) {
      sc[j] += c[j][i];
      if (sc[j]) t.a[j][0] = t.a[j][1] = t.a[j][2] = 0;
    }
    t = fpow(t, b[i] - b[i - 1]);
    ans = t * ans;
  }
  printf("%d", ans.a[1][0]);
  return 0;
}
