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 = 1000000007;
struct Matrix {
  long long a[3][3];
  Matrix() {
    memset(a, 0, sizeof(a));
    for (int i = 0; i < 3; i++) {
      a[i][i] = 1;
    }
  }
};
inline Matrix getMatrix(int mask) {
  Matrix ret;
  ret.a[0][1] = ret.a[1][0] = 1;
  ret.a[1][2] = ret.a[2][1] = 1;
  for (int i = 0; i < 3; i++) {
    if (mask & (1 << i)) {
      for (int j = 0; j < 3; j++) {
        ret.a[i][j] = 0;
      }
    }
  }
  return ret;
}
inline Matrix product(Matrix &a, Matrix &b) {
  Matrix ret = getMatrix(7);
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      for (int k = 0; k < 3; k++) {
        ret.a[i][j] += a.a[i][k] * b.a[k][j];
      }
      ret.a[i][j] %= mod;
    }
  }
  return ret;
}
inline Matrix power(Matrix A, long long n) {
  Matrix ret;
  while (n) {
    if (n & 1) ret = product(ret, A);
    A = product(A, A);
    n >>= 1;
  }
  return ret;
}
inline vector<pair<long long, long long> > compress(
    vector<pair<long long, long long> > a) {
  vector<pair<long long, long long> > ret;
  if (a.empty()) return ret;
  sort(a.begin(), a.end());
  long long l = a[0].first, r = a[0].second;
  for (auto x : a) {
    if (r + 1 < x.first) {
      ret.push_back({l, r});
      l = x.first;
      r = x.second;
    } else if (r < x.second) {
      r = x.second;
    }
  }
  ret.push_back({l, r});
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  int n;
  long long m;
  cin >> n >> m;
  vector<pair<long long, long long> > rw[3];
  int a;
  long long l, r;
  for (int i = 0; i < n; i++) {
    cin >> a >> l >> r;
    a--;
    rw[a].push_back({l, r});
  }
  map<long long, int> mp;
  mp[2] = 0;
  for (int i = 0; i < 3; i++) {
    rw[i] = compress(rw[i]);
    for (auto x : rw[i]) {
      mp[x.first] ^= (1 << i);
      mp[x.second + 1] ^= (1 << i);
    }
  }
  vector<pair<long long, int> > mask;
  int cur = 0;
  for (auto x : mp) {
    cur ^= x.second;
    mask.push_back({x.first, cur});
  }
  int sz = mask.size();
  Matrix ret;
  for (int i = 0; i < sz; i++) {
    Matrix A = getMatrix(mask[i].second);
    long long len = (i + 1 == sz) ? (m - mask[i].first + 1)
                                  : (mask[i + 1].first - mask[i].first);
    A = power(A, len);
    ret = product(A, ret);
  }
  cout << ret.a[1][1] << endl;
  return 0;
}
