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;
long long n, m, a[10005], l[10005], r[10005], num[4][20005];
vector<long long> p;
struct ma {
  long long c[4][4];
  ma() { memset(c, 0, sizeof c); }
  ma operator*(const ma& rem) const {
    ma res;
    for (long long i = 1; i <= 3; ++i) {
      for (long long j = 1; j <= 3; ++j) {
        for (long long k = 1; k <= 3; ++k) {
          res.c[i][j] += c[i][k] * rem.c[k][j];
          res.c[i][j] %= mod;
        }
      }
    }
    return res;
  }
} ans;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  p.push_back(1);
  p.push_back(m);
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i] >> l[i] >> r[i];
    p.push_back(l[i]);
    p.push_back(r[i] + 1);
  }
  sort(p.begin(), p.end());
  vector<long long>::iterator it = unique(p.begin(), p.end());
  p.resize(distance(p.begin(), it));
  for (long long i = 1; i <= n; ++i) {
    long long pos1 = lower_bound(p.begin(), p.end(), l[i]) - p.begin();
    long long pos2 = lower_bound(p.begin(), p.end(), r[i] + 1) - p.begin();
    num[a[i]][pos1]++;
    num[a[i]][pos2]--;
  }
  for (long long i = 1; i < p.size(); ++i) {
    for (long long j = 1; j <= 3; ++j) {
      num[j][i] += num[j][i - 1];
    }
  }
  ans.c[1][2] = 1;
  for (long long i = 1; i < p.size(); ++i) {
    long long leng = p[i] - p[i - 1];
    ma temp;
    if (!num[1][i - 1]) {
      temp.c[1][1] = temp.c[1][2] = 1;
    }
    if (!num[2][i - 1]) {
      temp.c[2][1] = temp.c[2][2] = temp.c[2][3] = 1;
    }
    if (!num[3][i - 1]) {
      temp.c[3][2] = temp.c[3][3] = 1;
    }
    for (; leng; leng >>= 1, temp = temp * temp)
      if (leng & 1) ans = ans * temp;
  }
  cout << ans.c[1][2];
  return 0;
}
