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 = 10000 + 10, MOD = int(1e9 + 7);
struct matrix {
  int d[3][3];
  matrix() { memset(d, 0, sizeof d); }
  int *operator[](int i) { return d[i]; }
};
matrix multiply(matrix a, matrix b) {
  matrix c;
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
      for (int k = 0; k < 3; k++) {
        c[i][k] = (c[i][k] + 1LL * a[i][j] * b[j][k]) % MOD;
      }
  return c;
}
matrix power(matrix base, matrix a, long long n) {
  matrix c = base;
  while (n > 0) {
    if (n & 1) c = multiply(c, a);
    a = multiply(a, a);
    n /= 2;
  }
  return c;
}
int n;
long long m;
map<long long, vector<pair<int, int> > > events;
int state[3];
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int a;
    long long l, r;
    cin >> a >> l >> r;
    a--;
    events[l - 1].push_back(make_pair(1, a));
    events[r].push_back(make_pair(-1, a));
  }
  events[m] = vector<pair<int, int> >();
  matrix res, tran;
  res[0][1] = 1;
  memset(state, 0, sizeof state);
  long long last = 1;
  for (auto &pp : events) {
    long long t = pp.first;
    if (t > last) {
      for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++)
          tran[i][j] = (state[j] > 0) ? 0 : (abs(i - j) <= 1);
      }
      res = power(res, tran, t - last);
    }
    last = t;
    sort(pp.second.begin(), pp.second.end());
    for (auto p : pp.second) {
      state[p.second] += p.first;
    }
  }
  printf("%d\n", res[0][1]);
  return 0;
}
