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 MAXN = 2e4 + 10;
const long long mod = 1e9 + 7;
struct obst {
  long long ind;
  long long w, hm;
} o[MAXN];
long long on;
bool comp(obst a, obst b) {
  if (a.ind != b.ind) return a.ind < b.ind;
  return a.w < b.w;
}
long long n, m;
void mult(long long a[3][3], long long b[3][3], long long c[3][3]) {
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      c[i][j] = 0;
      for (int k = 0; k < 3; k++) {
        c[i][j] += a[i][k] * b[k][j];
      }
      c[i][j] %= mod;
    }
  }
}
void pow(long long a[3][3], long long p, long long c[3][3]) {
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++) c[i][j] = (i == j);
  long long tmp[3][3];
  while (p) {
    if (p % 2) {
      mult(c, a, tmp);
      for (int i = 0; i < 3; i++)
        for (int j = 0; j < 3; j++) c[i][j] = tmp[i][j];
    }
    p /= 2;
    mult(a, a, tmp);
    for (int i = 0; i < 3; i++)
      for (int j = 0; j < 3; j++) a[i][j] = tmp[i][j];
  }
}
int main() {
  scanf(" %lld %lld", &n, &m);
  for (int i = 0; i < n; i++) {
    long long a, l, r;
    scanf(" %lld %lld %lld", &a, &l, &r);
    o[on].ind = l, o[on].w = a - 1, o[on++].hm = 1;
    o[on].ind = r + 1, o[on].w = a - 1, o[on++].hm = -1;
  }
  sort(o, o + on, comp);
  long long a[3][3] = {{0, 1, 0}, {0, 0, 0}, {0, 0, 0}};
  long long pos = 1;
  long long base[3][3] = {{1, 1, 0}, {1, 1, 1}, {0, 1, 1}};
  long long tmp[3][3];
  long long use[3] = {0, 0, 0};
  for (int it = 0; it < on; it++) {
    long long cupo[3][3];
    for (int i = 0; i < 3; i++)
      for (int j = 0; j < 3; j++) cupo[i][j] = (use[j] == 0 ? base[i][j] : 0);
    pow(cupo, o[it].ind - pos - 1, tmp);
    for (int i = 0; i < 3; i++)
      for (int j = 0; j < 3; j++) cupo[i][j] = tmp[i][j];
    mult(a, cupo, tmp);
    for (int i = 0; i < 3; i++)
      for (int j = 0; j < 3; j++) a[i][j] = tmp[i][j];
    while (it + 1 < on && o[it + 1].ind == o[it].ind) {
      use[o[it].w] += o[it].hm;
      it++;
    }
    use[o[it].w] += o[it].hm;
    pos = o[it].ind - 1;
  }
  pow(base, m - pos, tmp);
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++) base[i][j] = tmp[i][j];
  mult(a, base, tmp);
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++) a[i][j] = tmp[i][j];
  printf("%lld\n", a[0][1]);
}
