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 int mod = 1e9 + 7;
template <int maxn>
struct Matrix {
  long long int entry[maxn][maxn];
  Matrix() { memset(entry, 0, sizeof entry); }
  Matrix<maxn> operator*(const Matrix<maxn> &rhs) const {
    Matrix<maxn> res;
    for (int i = 0; i < maxn; i++)
      for (int j = 0; j < maxn; j++)
        for (int k = 0; k < maxn; k++)
          res.entry[i][j] =
              (res.entry[i][j] + entry[i][k] * rhs.entry[k][j]) % mod;
    return res;
  }
};
int cnt[3];
vector<pair<pair<long long int, int>, int> > v;
Matrix<3> exp(Matrix<3> x, long long int y) {
  if (y == 1) return x;
  Matrix<3> tmp = exp(x, y / 2);
  tmp = tmp * tmp;
  if (y % 2)
    return tmp * x;
  else
    return tmp;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int n, x, l, r, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> x >> l >> r;
    v.push_back({{l, x - 1}, 1});
    v.push_back({{r + 1, x - 1}, -1});
  }
  sort(v.begin(), v.end());
  int idx = 0;
  long long int now = 1, pre = 1;
  Matrix<3> trans, init;
  init.entry[0][0] = init.entry[1][1] = init.entry[2][2] = 1;
  trans.entry[0][0] = 1;
  trans.entry[0][1] = 1;
  trans.entry[0][2] = 0;
  trans.entry[1][0] = 1;
  trans.entry[1][1] = 1;
  trans.entry[1][2] = 1;
  trans.entry[2][0] = 0;
  trans.entry[2][1] = 1;
  trans.entry[2][2] = 1;
  while (idx < v.size()) {
    now = v[idx].first.first;
    init = exp(trans, now - pre) * init;
    cnt[v[idx].first.second] += v[idx].second;
    while (idx + 1 < v.size() && v[idx + 1].first.first == now) {
      idx++;
      cnt[v[idx].first.second] += v[idx].second;
    }
    if (!cnt[0]) {
      trans.entry[0][0] = 1;
      trans.entry[1][0] = 1;
      trans.entry[2][0] = 0;
    } else {
      trans.entry[0][0] = 0;
      trans.entry[1][0] = 0;
      trans.entry[2][0] = 0;
    }
    if (!cnt[1]) {
      trans.entry[0][1] = 1;
      trans.entry[1][1] = 1;
      trans.entry[2][1] = 1;
    } else {
      trans.entry[0][1] = 0;
      trans.entry[1][1] = 0;
      trans.entry[2][1] = 0;
    }
    if (!cnt[2]) {
      trans.entry[0][2] = 0;
      trans.entry[1][2] = 1;
      trans.entry[2][2] = 1;
    } else {
      trans.entry[0][2] = 0;
      trans.entry[1][2] = 0;
      trans.entry[2][2] = 0;
    }
    pre = now;
    idx++;
  }
  if (now != m) init = exp(trans, m - pre) * init;
  cout << init.entry[1][1] << endl;
}
