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;
const long long N = 3e5 + 5;
const long long INF = 1e14;
map<long long, vector<pair<long long, long long> > > idx;
long long ret[3][3];
long long res1[3];
void matmul(long long A[3][3], long long C[3][3]) {
  for (long long i = 0; i < 3; i++) {
    for (long long j = 0; j < 3; j++) {
      ret[i][j] = 0;
      for (long long k = 0; k < 3; k++) {
        ret[i][j] = (ret[i][j] + (A[i][k] * C[k][j]) % MOD) % MOD;
      }
    }
  }
}
void mpow(long long A[3][3], long long p) {
  if (p == 0) {
    for (long long i = 0; i < 3; i++)
      for (long long j = 0; j < 3; j++) ret[i][j] = 0;
    ret[0][0] = ret[1][1] = ret[2][2] = 1;
    return;
  }
  if (p == 1) {
    for (long long i = 0; i < 3; i++)
      for (long long j = 0; j < 3; j++) ret[i][j] = A[i][j];
    return;
  }
  long long temp[3][3];
  mpow(A, p / 2);
  for (long long i = 0; i < 3; i++)
    for (long long j = 0; j < 3; j++) temp[i][j] = ret[i][j];
  matmul(temp, temp);
  for (long long i = 0; i < 3; i++)
    for (long long j = 0; j < 3; j++) temp[i][j] = ret[i][j];
  if (p % 2 == 1) matmul(temp, A);
}
void vecmul(long long A[3][3], long long b[3]) {
  for (long long i = 0; i < 3; i++) {
    res1[i] = 0;
    for (long long j = 0; j < 3; j++) {
      res1[i] = (res1[i] + (A[i][j] * b[j]) % MOD) % MOD;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    long long a, lf, rt;
    cin >> a >> lf >> rt;
    idx[lf].push_back(make_pair(a, 0));
    idx[rt + 1].push_back(make_pair(a, 1));
  }
  int a[4] = {0, 0, 0, 0};
  long long ans[3];
  ans[0] = 0;
  ans[1] = 1;
  ans[2] = 0;
  long long A[3][3] = {{1, 1, 0}, {1, 1, 1}, {0, 1, 1}};
  long long B[3][3];
  for (long long i = 0; i < 3; i++)
    for (long long j = 0; j < 3; j++) B[i][j] = A[i][j];
  long long prev = 1;
  for (auto it : idx) {
    long long cur = it.first;
    long long p = cur - prev;
    mpow(B, p);
    vecmul(ret, ans);
    for (long long i = 0; i < 3; i++) ans[i] = res1[i];
    if (0) {
      cerr << "Multiplying with \n";
      for (long long i = 0; i < 3; i++) {
        for (long long j = 0; j < 3; j++) cerr << B[i][j] << " ";
        cerr << "\n";
      }
      cerr << "to the power " << p << "\n";
      for (long long i = 0; i < 3; i++) cerr << ans[i] << " ";
      cerr << "\n";
    }
    for (int i = 0; i < it.second.size(); ++i) {
      if (it.second[i].second == 1) a[it.second[i].first]--;
    }
    for (int i = 0; i < it.second.size(); ++i) {
      if (it.second[i].second == 0) a[it.second[i].first]++;
    }
    for (int i = 0; i < 3; ++i) {
      for (long long j = 0; j < 3; j++) {
        B[i][j] = A[i][j];
        if (a[j + 1] >= 1) {
          B[i][j] = 0;
          ans[j] = 0;
        }
      }
    }
    prev = cur;
  }
  long long p = m - prev;
  mpow(B, p);
  vecmul(ret, ans);
  for (long long i = 0; i < 3; i++) ans[i] = res1[i];
  cout << ans[1] << "\n";
  return 0;
}
