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 = 1e5 + 5;
const long long mod = 1000000007;
int n, sl = 0;
long long m, a[maxn], l[maxn], r[maxn], seg[maxn];
long long mul[maxn][3];
struct matrix {
  long long cur[3][3];
  matrix() {
    for (int i = 0; i <= 2; ++i)
      for (int j = 0; j <= 2; ++j) cur[i][j] = 0;
  }
};
matrix operator*(matrix a, matrix b) {
  matrix c;
  for (int i = 0; i <= 2; ++i)
    for (int j = 0; j <= 2; ++j)
      for (int k = 0; k <= 2; ++k)
        c.cur[i][j] = (c.cur[i][j] + a.cur[i][k] * b.cur[k][j]) % mod;
  return c;
}
matrix f;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i] >> l[i] >> r[i];
    seg[++sl] = l[i];
    seg[++sl] = r[i] + 1;
  }
  seg[++sl] = 2;
  seg[++sl] = m + 1;
  sort(seg + 1, seg + sl + 1);
  sl = unique(seg + 1, seg + sl + 1) - seg - 1;
  for (int i = 1; i <= n; ++i) {
    l[i] = lower_bound(seg + 1, seg + sl + 1, l[i]) - seg;
    r[i] = lower_bound(seg + 1, seg + sl + 1, r[i] + 1) - seg;
    mul[l[i]][a[i] - 1]++;
    mul[r[i]][a[i] - 1]--;
  }
  for (int i = 0; i <= 2; ++i) f.cur[i][i] = 1;
  for (int i = 1; i < sl; ++i) {
    for (int j = 0; j <= 2; ++j) mul[0][j] += mul[i][j];
    matrix g;
    for (int j = 0; j <= 2; ++j)
      for (int k = max(0, j - 1); k <= min(2, j + 1); ++k)
        if (mul[0][k] == 0) g.cur[j][k] = 1;
    long long length = seg[i + 1] - seg[i];
    for (; length > 0; length >>= 1, g = g * g)
      if ((length & 1) == 1) f = f * g;
  }
  cout << f.cur[1][1];
}
