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>
int t, n, x[20005], b[3], d[20005][3], num;
long long a[20005], L[20005], R[20005], m;
const int mo = 1e9 + 7;
long long read() {
  long long ans = 0;
  char c = getchar();
  while (c > '9' || c < '0') c = getchar();
  while (c <= '9' && c >= '0') ans = ans * 10 + c - '0', c = getchar();
  return ans;
}
int find(long long x) {
  int l = 1, r = t;
  while (l <= r) {
    int mid = l + r >> 1;
    if (a[mid] == x) return mid;
    if (a[mid] < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
}
struct M {
  int v[3][3];
};
M q = {{{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}};
M operator*(M a, M b) {
  M ans = {{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}};
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
      for (int k = 0; k < 3; k++)
        (ans.v[i][j] += 1ll * a.v[i][k] * b.v[k][j] % mo) %= mo;
  return ans;
}
M js(M a, long long k) {
  M ans = {{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}};
  while (k) {
    if (k & 1) ans = ans * a;
    a = a * a;
    k >>= 1;
  }
  return ans;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    x[i] = read();
    L[i] = read();
    R[i] = read();
    L[i]--;
    x[i]--;
    a[++num] = L[i], a[++num] = R[i];
  }
  a[++num] = m;
  a[++num] = 1;
  t = 1;
  std::sort(a + 1, a + 1 + num);
  for (int i = 2; i <= num; i++)
    if (a[t] != a[i]) a[++t] = a[i];
  for (int i = 1; i <= n; i++) d[find(L[i])][x[i]]++, d[find(R[i])][x[i]]--;
  M s = {{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}};
  for (int i = 1; i < t; i++) {
    M now = q;
    for (int j = 0; j < 3; j++) {
      b[j] += d[i][j];
      if (b[j]) now.v[0][j] = now.v[1][j] = now.v[2][j] = 0;
    }
    s = s * js(now, a[i + 1] - a[i]);
  }
  int ans = 0;
  printf("%d\n", s.v[1][1]);
}
