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 long long MOD = 1e9 + 7;
struct matrix {
  long long t[3][3];
  matrix(int k) {
    if (k == -1) {
      for (int i = 0; i < (3); ++i)
        for (int j = 0; j < (3); ++j) t[i][j] = 0;
    } else {
      for (int i = 0; i < (3); ++i)
        for (int j = 0; j < (3); ++j) t[i][j] = 1;
      t[0][2] = t[2][0] = 0;
      for (int i = 0; i < (3); ++i)
        if ((((k) >> (i)) & 1))
          for (int j = 0; j < (3); ++j) t[i][j] = 0;
    }
  }
};
matrix operator*(matrix a, matrix b) {
  matrix c(-1);
  for (int i = 0; i < (3); ++i)
    for (int j = 0; j < (3); ++j)
      for (int k = 0; k < (3); ++k) {
        c.t[i][j] = (c.t[i][j] + a.t[i][k] * b.t[k][j] % MOD) % MOD;
      }
  return c;
}
matrix POW(matrix a, long long k) {
  if (k == 1) return a;
  matrix c = POW(a, k >> 1ll);
  c = c * c;
  if (k % 2) c = c * a;
  return c;
}
int cnt[4];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, int> > ban;
  for (int i = (1); i <= (n); ++i) {
    long long a, l, r;
    cin >> a >> l >> r;
    ban.push_back(pair<long long, int>(l, a));
    if (r != m) ban.push_back(pair<long long, int>(r + 1, -a));
  }
  sort(ban.begin(), ban.end());
  matrix ans(-1);
  ans.t[1][0] = 1;
  long long col = 1, status = 0;
  ban.push_back(pair<long long, int>(m + 1, 4));
  for (pair<long long, int> x : ban) {
    if (x.first != col) {
      matrix now(status);
      if (col != 1)
        ans = POW(now, x.first - col) * ans;
      else {
        if (x.first - col > 1) ans = POW(now, x.first - col - 1) * ans;
      }
      col = x.first;
    }
    if (x.first == 1) {
      if (x.second == -2) {
        cout << 0;
        return 0;
      }
    }
    if (x.second < 0) {
      cnt[-x.second]--;
      if (cnt[-x.second] == 0) status ^= (1 << (-x.second - 1));
    } else {
      if (cnt[x.second] == 0) status ^= (1 << (x.second - 1));
      cnt[x.second]++;
    }
  }
  cout << ans.t[1][0];
}
