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;
vector<vector<long long> > operator*(vector<vector<long long> > a,
                                     vector<vector<long long> > b) {
  int n = (int)a.size();
  int m = (int)b.size();
  int p = (int)b[0].size();
  vector<vector<long long> > c;
  c.resize(n);
  for (int i = 0; i < n; i++) {
    c[i].assign(p, 0);
    for (int j = 0; j < p; j++) {
      for (int k = 0; k < m; k++) {
        c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % 1000000007;
      }
    }
  }
  return c;
}
vector<vector<long long> > id(int size) {
  vector<vector<long long> > a(size, vector<long long>(size, 0));
  for (int i = 0; i < size; i++) a[i][i] = 1;
  return a;
}
vector<vector<long long> > matrixExp(vector<vector<long long> > a,
                                     long long n) {
  if (n == 0) return id(a.size());
  vector<vector<long long> > c = matrixExp(a, n / 2);
  c = c * c;
  if (n % 2 != 0) c = c * a;
  return c;
}
vector<vector<long long> > blocked(int *b) {
  vector<vector<long long> > a(3, vector<long long>(3, 0));
  a[0][0] = a[0][1] = 1;
  a[1][0] = a[1][1] = a[1][2] = 1;
  a[2][1] = a[2][2] = 1;
  if (b[1] > 0) a[0][0] = a[0][1] = a[1][0] = 0;
  if (b[2] > 0) a[0][1] = a[1][0] = a[1][1] = a[1][2] = a[2][1] = 0;
  if (b[3] > 0) a[1][2] = a[2][1] = a[2][2] = 0;
  return a;
}
int main() {
  int n;
  long long m;
  cin >> n >> m;
  vector<pair<long long, int> > block;
  for (int i = 0; i < n; i++) {
    int b;
    long long l, r;
    cin >> b >> l >> r;
    block.push_back(pair<long long, int>(l, b));
    block.push_back(pair<long long, int>(r, -b));
  }
  sort(block.begin(), block.end());
  vector<vector<long long> > ans = id(3);
  long long pos = 1;
  int b[4] = {0, 0, 0, 0};
  for (auto it : block) {
    long long nextPos = it.first;
    ans = ans * matrixExp(blocked(b), nextPos - pos);
    b[abs(it.second)] += it.second;
    pos = nextPos;
  }
  ans = ans * matrixExp(blocked(b), m - pos);
  cout << ans[1][1] << endl;
}
