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;
int mo = 1e9 + 7;
struct r {
  long long s[5][5];
  r() { memset(s, 0, sizeof s); }
  void cr() {
    s[1][1] = s[1][2] = s[2][1] = s[2][2] = s[2][3] = s[3][2] = s[3][3] = 1;
    s[1][3] = s[3][1] = 0;
  }
} tmp, ans;
int n, t, c[5][40050], pre[5];
struct p {
  int a;
  long long l, r;
} a[10050];
long long m, s[20050];
r operator*(r a, r b) {
  r ret;
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++)
      for (int k = 1; k <= 3; k++) {
        ret.s[i][j] += a.s[i][k] * b.s[k][j], ret.s[i][j] %= mo;
      }
  return ret;
}
void ksm(long long t) {
  r p;
  p = tmp;
  while (t) {
    if (t & 1) tmp = tmp * p;
    p = p * p;
    t >>= 1;
  }
}
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);
    s[++t] = a[i].l - 1;
    s[++t] = a[i].r;
  }
  s[++t] = 1;
  s[++t] = m;
  sort(s + 1, s + t + 1);
  t = unique(s + 1, s + t + 1) - s - 1;
  for (int i = 1; i <= n; i++) {
    int l = lower_bound(s + 1, s + t + 1, a[i].l - 1) - s;
    int r = lower_bound(s + 1, s + t + 1, a[i].r) - s;
    c[a[i].a][l]++;
    c[a[i].a][r]--;
  }
  ans.s[1][1] = ans.s[1][3] = 0;
  ans.s[1][2] = 1;
  for (int i = 1; i < t; i++) {
    tmp.cr();
    for (int j = 1; j <= 3; j++) {
      pre[j] += c[j][i];
      if (pre[j]) tmp.s[1][j] = tmp.s[2][j] = tmp.s[3][j] = 0;
    }
    ksm(s[i + 1] - s[i] - 1);
    ans = ans * tmp;
  }
  printf("%lld", ans.s[1][2]);
}
