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;
long long n = 3, MOD = 1e9 + 7;
vector<vector<long long> > matMul(vector<vector<long long> > &a,
                                  vector<vector<long long> > &b) {
  vector<vector<long long> > c(n, vector<long long>(n, 0));
  int k;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (k = 0; k < n; k++) {
        c[i][j] += (a[i][k] % MOD) * (b[k][j] % MOD);
        c[i][j] %= MOD;
      }
  return c;
}
vector<vector<long long> > matPow(vector<vector<long long> > base,
                                  long long p) {
  vector<vector<long long> > ans(n, vector<long long>(n, 0));
  int i, j;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) ans[i][j] = (i == j);
  while (p) {
    if (p & 1) ans = matMul(ans, base);
    base = matMul(base, base);
    p >>= 1;
  }
  return ans;
}
vector<vector<long long> > org = {{1, 1, 0}, {1, 1, 1}, {0, 1, 1}};
vector<long long> block = {0, 0, 0};
int main() {
  ios::sync_with_stdio(0);
  long long o, m;
  cin >> o >> m;
  vector<pair<pair<long long, long long>, long long> > obs;
  for (int i = 0; i < o; i++) {
    long long a, l, r;
    cin >> a >> l >> r;
    obs.push_back(pair<pair<long long, long long>, long long>(
        pair<long long, long long>(l - 1, -1), a));
    obs.push_back(pair<pair<long long, long long>, long long>(
        pair<long long, long long>(r, 1), a));
  }
  vector<long long> isBlc = {0, 0, 0};
  vector<vector<long long> > ans(3, vector<long long>(3, 0));
  ans[0] = {0, 1, 0};
  long long cur = 1;
  sort(obs.begin(), obs.end());
  for (auto &e : obs) {
    if (e.first.first < cur) {
      isBlc[e.second - 1] -= e.first.second;
      continue;
    }
    vector<vector<long long> > curMat(3, vector<long long>(3));
    for (int i = 0; i < 3; i++) {
      vector<long long> toc;
      if (isBlc[i])
        toc = block;
      else
        toc = org[i];
      for (int j = 0; j < 3; j++) curMat[j][i] = toc[j];
    }
    vector<vector<long long> > c =
        matPow(curMat, e.first.first - cur + (cur > 1));
    ans = matMul(ans, c);
    cur = e.first.first + 1;
    isBlc[e.second - 1] -= e.first.second;
  }
  vector<vector<long long> > curMat(3, vector<long long>(3));
  for (int i = 0; i < 3; i++) {
    vector<long long> toc;
    if (isBlc[i])
      toc = block;
    else
      toc = org[i];
    for (int j = 0; j < 3; j++) curMat[j][i] = toc[j];
  }
  vector<vector<long long> > c = matPow(curMat, m - cur + 1);
  ans = matMul(ans, c);
  cout << ans[0][1] << endl;
  return 0;
}
