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 = 10101, P = 1e9 + 7;
int n, cntn, cnt;
int blo[N << 1][3];
long long m;
long long tim[N << 1], L[N << 1][3], R[N << 1][3];
struct NN {
  int x, ty;
  long long y;
  bool operator<(const NN &c) const { return y < c.y; }
} nn[N << 1];
void upd(long long &a, long long b) {
  a += b;
  if (a >= P) a -= P;
}
struct M {
  long long a[3][3];
  M() { memset(a, 0, sizeof(a)); }
  void E() {
    for (int i = (0); i < (3); i++) a[i][i] = 1;
  }
  M operator*(const M &c) const {
    M res;
    for (int i = (0); i < (3); i++)
      for (int j = (0); j < (3); j++)
        for (int k = (0); k < (3); k++) {
          upd(res.a[i][j], a[i][k] * c.a[k][j] % P);
        }
    return res;
  }
};
M kpow(M a, long long b) {
  M res;
  res.E();
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
void getR(int c) {
  M a;
  for (int i = (0); i < (3); i++)
    for (int j = (0); j < (3); j++) {
      if (abs(i - j) <= 1 && !blo[c][i] && !blo[c][j]) a.a[i][j] = 1;
    }
  a = kpow(a, tim[c + 1] - 1 - tim[c]);
  for (int i = (0); i < (3); i++) {
    for (int j = (0); j < (3); j++) upd(R[c][i], a.a[i][j] * L[c][j] % P);
  }
}
int main() {
  while (~scanf("%d%lld", &n, &m)) {
    cntn = cnt = 0;
    memset(blo, 0, sizeof(blo));
    memset(L, 0, sizeof(L));
    memset(R, 0, sizeof(L));
    for (int i = (1); i < (n + 1); i++) {
      int a;
      long long l, r;
      scanf("%d%lld%lld", &a, &l, &r);
      nn[++cntn] = NN{a - 1, 1, l};
      nn[++cntn] = NN{a - 1, -1, r + 1};
    }
    sort(nn + 1, nn + cntn + 1);
    tim[++cnt] = 1;
    for (int i = (1); i < (cntn + 1); i++) {
      if (nn[i].y != tim[cnt]) {
        tim[++cnt] = nn[i].y;
        for (int j = (0); j < (3); j++) blo[cnt][j] = blo[cnt - 1][j];
      }
      blo[cnt][nn[i].x] += nn[i].ty;
    }
    tim[cnt + 1] = m + 1;
    L[1][1] = 1;
    getR(1);
    for (int c = (2); c < (cnt + 1); c++) {
      for (int i = (0); i < (3); i++)
        for (int j = (0); j < (3); j++)
          if (abs(i - j) <= 1 && !blo[c][i] && !blo[c - 1][j])
            upd(L[c][i], R[c - 1][j]);
      getR(c);
    }
    printf("%lld\n", R[cnt][1]);
  }
  return 0;
}
