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;
void READ() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
}
const int N = 5 + 1e4;
const int M = 1e9 + 7;
long long n, m, T, ans, ma, s, t;
const int rnk = 3;
struct Matrix {
  long long a[rnk][rnk];
  Matrix() { memset(a, 0, sizeof(a)); }
  Matrix operator*(const Matrix& t) const {
    Matrix ans;
    for (int i = 0; i < rnk; i++)
      for (int j = 0; j < rnk; j++)
        for (int k = 0; k < rnk; k++)
          ans.a[i][j] = (ans.a[i][j] + a[i][k] * t.a[k][j]) % M;
    return ans;
  }
} an, mt, mt0;
Matrix Pow(Matrix a, long long p) {
  Matrix ans;
  for (int i = 0; i < rnk; i++) {
    ans.a[i][i] = 1;
  }
  while (p) {
    if (p & 1) ans = ans * a;
    a = a * a;
    p >>= 1;
  }
  return ans;
}
void init() {
  for (int i = 0; i < rnk; i++) {
    an.a[i][i] = 1;
  }
  for (int i = 0; i < rnk; i++) {
    for (int j = 0; j < rnk; j++) {
      mt.a[i][j] = mt0.a[i][j] = 1;
    }
  }
  mt.a[rnk - 1][0] = mt0.a[rnk - 1][0] = 0;
  mt.a[0][rnk - 1] = mt0.a[0][rnk - 1] = 0;
}
void upd(int j, int f) {
  if (f == 1)
    for (int i = 0; i < rnk; i++) mt.a[i][j] = 0;
  else if (f == 0) {
    for (int i = 0; i < rnk; i++) mt.a[i][j] = mt0.a[i][j];
  }
}
struct Node {
  long long i, j, f;
  bool operator<(const Node& t) const { return j < t.j; }
} b[2 * N];
int num[4];
int main() {
  READ();
  cin >> n >> m;
  long long a, l, r, k = 0;
  for (int i = 0; i < n; i++) {
    cin >> a >> l >> r;
    b[k++] = Node{a, l, 1};
    b[k++] = Node{a, r + 1, 0};
  }
  sort(b, b + k);
  long long laj = 2;
  init();
  for (int i = 0; i < k; i++) {
    int a = b[i].i - 1;
    if (b[i].f == 1)
      num[a]++;
    else
      num[a]--;
    an = an * Pow(mt, b[i].j - laj);
    upd(a, num[a]);
    laj = b[i].j;
  }
  an = an * Pow(mt, m - laj);
  ans = 0;
  for (int i = 0; i < rnk; i++) {
    ans += an.a[1][i];
  }
  cout << ans % M;
  return 0;
}
