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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
const long long mod = 1e9 + 7;
class matrix {
 public:
  vector<vector<long long> > v;
  matrix() { v = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}; }
  vector<long long>& operator[](long long loc) { return v[loc]; }
};
matrix mult(matrix a, matrix b) {
  matrix c;
  for (long long i = 0; i < 3; i++)
    for (long long j = 0; j < 3; j++) {
      c.v[i][j] = 0;
      for (long long k = 0; k < 3; k++)
        c.v[i][j] = (c.v[i][j] + a.v[i][k] * b.v[k][j]) % mod;
    }
  return c;
}
matrix power(matrix base, long long expo) {
  matrix ans;
  while (expo) {
    if (expo & 1) {
      ans = mult(base, ans);
    }
    base = mult(base, base);
    expo = expo >> 1;
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> b(4, 0);
  vector<pair<pair<long long, long long>, long long> > v(n);
  multiset<pair<long long, long long> > s;
  for (long long i = 0; i < n; i++) {
    cin >> v[i].second >> v[i].first.first >> v[i].first.second;
    s.insert({v[i].first.first, v[i].second});
    s.insert({v[i].first.second + 1, -v[i].second});
  }
  long long prev = 1;
  matrix ans, c;
  c[0] = {1, 1, 0};
  c[1] = {1, 1, 1};
  c[2] = {0, 1, 1};
  for (auto& i : s) {
    long long loc = abs(i.second);
    long long curr = i.first - 1;
    if (curr > prev) {
      ans = mult(power(c, curr - prev), ans);
      prev = curr;
    }
    long long val = i.second / loc;
    loc--;
    b[loc] += val;
    if (b[loc] <= 0) {
      c[loc] = {loc != 2, 1, loc != 0};
    } else {
      c[loc] = {0, 0, 0};
    }
  }
  ans = mult(power(c, m - prev), ans);
  cout << ans[1][1];
  return 0;
}
