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 sz = 3;
const long long mod = 1e9 + 7;
struct matrix {
  long long mat[sz][sz];
  void inti() { memset(mat, 0, sizeof mat); }
  matrix operator*(matrix b) {
    matrix c;
    c.inti();
    for (int i = 0; i < sz; ++i) {
      for (int j = 0; j < sz; ++j) {
        for (int k = 0; k < sz; ++k)
          c.mat[i][j] = (c.mat[i][j] + (mat[i][k] * b.mat[k][j]) % mod) % mod;
      }
    }
    return c;
  }
} base[8];
matrix matrix_pow(matrix a, long long n) {
  matrix ans;
  ans.inti();
  for (int i = 0; i < sz; ++i) ans.mat[i][i] = 1;
  while (n) {
    if (n & 1) ans = ans * a;
    a = a * a;
    n /= 2;
  }
  return ans;
}
struct node {
  int a, v;
  long long s;
  node(int a = 0, int b = 0, long long c = 0) : a(a), v(b), s(c){};
  bool operator<(const node &b) const { return s < b.s; }
};
vector<node> v;
int cnt[4];
int cal() { return (cnt[1] ? 4 : 0) + (cnt[2] ? 2 : 0) + (cnt[3] ? 1 : 0); }
void print(matrix a) {
  for (int i = 0; i < sz; ++i) {
    for (int j = 0; j < sz; ++j) cout << a.mat[i][j] << " ";
    cout << endl;
  }
  cout << endl;
}
int main() {
  int n;
  long long m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    int a;
    long long l, r;
    cin >> a >> l >> r;
    v.push_back(node(a, 1, l - 1));
    v.push_back(node(a, -1, r));
  }
  v.push_back(node(0, 0, m));
  sort(v.begin(), v.end());
  long long pre = 1;
  for (int i = 0; i < 8; ++i) base[i].inti();
  for (int i = 0; i < 8; ++i) {
    if (!(i & 4)) base[i].mat[0][0] = base[i].mat[1][0] = 1;
    if (!(i & 2)) base[i].mat[0][1] = base[i].mat[1][1] = base[i].mat[2][1] = 1;
    if (!(i & 1)) base[i].mat[1][2] = base[i].mat[2][2] = 1;
  }
  matrix ans;
  ans.inti();
  ans.mat[0][1] = 1;
  for (auto it : v) {
    ans = ans * matrix_pow(base[cal()], it.s - pre);
    pre = it.s;
    cnt[it.a] += it.v;
  }
  cout << ans.mat[0][1] << endl;
}
