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 SZ = 3;
const long long MOD = 1e9 + 7;
long long add(long long a, long long b) {
  long long res = a + b;
  if (res >= MOD) {
    return res - MOD;
  }
  return res;
}
long long mult(long long a, long long b) {
  long long res = a;
  res *= b;
  if (res >= MOD) {
    return res % MOD;
  }
  return res;
}
struct matrix {
  long long arr[SZ][SZ];
  void reset() { memset(arr, 0, sizeof(arr)); }
  void makeiden() {
    reset();
    for (long long i = 0; i < SZ; ++i) {
      arr[i][i] = 1;
    }
  }
  matrix operator+(const matrix &o) const {
    matrix res;
    for (long long i = 0; i < SZ; ++i) {
      for (long long j = 0; j < SZ; ++j) {
        res.arr[i][j] = add(arr[i][j], o.arr[i][j]);
      }
    }
    return res;
  }
  matrix operator*(const matrix &o) const {
    matrix res;
    for (long long i = 0; i < SZ; ++i) {
      for (long long j = 0; j < SZ; ++j) {
        res.arr[i][j] = 0;
        for (long long k = 0; k < SZ; ++k) {
          res.arr[i][j] = add(res.arr[i][j], mult(arr[i][k], o.arr[k][j]));
        }
      }
    }
    return res;
  }
};
matrix power(matrix a, long long b) {
  matrix res;
  res.makeiden();
  while (b) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long n, m;
long long block[SZ], res[SZ], res2[SZ];
map<long long, vector<long long> > events;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    long long a, l, r;
    cin >> a >> l >> r;
    events[l - 1].push_back(a);
    events[r].push_back(-a);
  }
  events[m].push_back(0);
  matrix ans;
  long long prev = 1;
  ans.makeiden();
  res[1] = 1;
  for (auto it : events) {
    long long cur = it.first;
    matrix mat;
    for (long long i = 0; i < SZ; i++) {
      for (long long j = 0; j < SZ; j++) {
        if (block[i] == 0 && abs(i - j) <= 1)
          mat.arr[i][j] = 1;
        else
          mat.arr[i][j] = 0;
      }
    }
    matrix expo = power(mat, cur - prev);
    ans = ans * expo;
    for (long long i = 0; i < SZ; i++) {
      res2[i] = 0;
      for (long long j = 0; j < SZ; j++)
        res2[i] = add(res2[i], mult(expo.arr[i][j], res[j]));
    }
    for (long long i = 0; i < SZ; i++) res[i] = res2[i];
    for (auto x : it.second) {
      long long a = abs(x);
      if (x < 0)
        block[a - 1]--;
      else
        block[a - 1]++;
    }
    prev = cur;
  }
  cout << res[1];
  return 0;
}
