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;
struct matrix {
  int n, m;
  int a[3][3];
  void init() { memset(a, 0, sizeof a); }
};
int mod = (1e9) + 7;
matrix T = {3, 3, {{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}};
matrix A = {1, 3, {{0, 1, 0}, {0, 0, 0}, {0, 0, 0}}};
matrix I = {3, 3, {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}};
int n;
long long m;
vector<pair<pair<long long, int>, int> > v;
matrix mul(matrix A, matrix B) {
  matrix C;
  C.n = A.n;
  C.m = B.m;
  C.init();
  for (int i = 0; i < A.n; ++i)
    for (int j = 0; j < B.m; ++j)
      for (int k = 0; k < A.m; ++k)
        C.a[i][j] = (C.a[i][j] + (1LL * A.a[i][k] * B.a[k][j]) % mod) % mod;
  return C;
}
matrix pow(matrix b, long long p) {
  if (p == 0) return I;
  matrix C = pow(b, p / 2);
  C = mul(C, C);
  if (p % 2) C = mul(C, b);
  return C;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    int a;
    long long l, r;
    cin >> a >> l >> r;
    v.push_back({{l, 0}, a});
    v.push_back({{r, 1}, a});
  }
  sort(v.begin(), v.end());
  v.push_back({{m, 1}, 0});
  long long lst = 1;
  int ch[3] = {0, 0, 0};
  for (int j = 0; j < v.size(); ++j) {
    long long in = v[j].first.first;
    int ty = v[j].first.second;
    int r = v[j].second - 1;
    long long stps = in - lst;
    matrix tmp = T;
    for (int i = 0; i < 3; ++i) {
      if (ch[i]) tmp.a[0][i] = tmp.a[1][i] = tmp.a[2][i] = 0;
    }
    A = mul(A, pow(tmp, stps));
    if (ty == 0) {
      ch[r]++;
      A.a[0][r] = 0;
    } else
      ch[r]--;
    lst = in;
  }
  cout << A.a[0][1] << endl;
  return 0;
}
