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 = 20000;
const long long MOD = 1e9 + 7;
long long a[3], b[3][3], c[3][3];
int n;
int sum[3];
long long m;
struct node {
  int x;
  long long wz;
  bool sig;
} A[N * 2];
bool cmp(const node& a, const node& b) { return a.wz < b.wz; }
void work(long long l, long long r) {
  memset(b, 0, sizeof(b));
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
      if (abs(i - j) <= 1 && !sum[j]) b[i][j] = 1;
  long long len = r - l;
  while (len) {
    if (len & 1) {
      for (int i = 0; i < 3; i++)
        for (int j = 0; j < 3; j++) (c[0][j] += a[i] * b[i][j] % MOD) %= MOD;
      for (int i = 0; i < 3; i++) a[i] = c[0][i], c[0][i] = 0;
    }
    for (int i = 0; i < 3; i++)
      for (int j = 0; j < 3; j++)
        for (int k = 0; k < 3; k++) (c[i][k] += b[i][j] * b[j][k] % MOD) %= MOD;
    for (int i = 0; i < 3; i++)
      for (int j = 0; j < 3; j++) b[i][j] = c[i][j], c[i][j] = 0;
    len /= 2;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    int x;
    long long l, r;
    cin >> x >> l >> r;
    A[i * 2 - 1].x = x - 1;
    A[i * 2 - 1].wz = l;
    A[i * 2 - 1].sig = 0;
    A[i * 2].x = x - 1;
    A[i * 2].wz = r + 1;
    A[i * 2].sig = 1;
  }
  sort(A + 1, A + n + n + 1, cmp);
  long long l = 1;
  a[1] = 1;
  for (int i = 1; i <= n + n; i++) {
    if (A[i].wz != A[i - 1].wz) {
      work(l, (long long)A[i].wz - 1);
      l = A[i].wz - 1;
    }
    if (A[i].sig)
      sum[A[i].x]--;
    else
      sum[A[i].x]++;
  }
  work(l, m);
  cout << a[1];
  return 0;
}
