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 inf = 1000000007;
vector<vector<long long> > dop;
void prin(vector<vector<long long> > a) {
  cout << endl << endl;
  for (long long c = 0; c < a.size(); c++) {
    for (long long i = 0; i < a[c].size(); i++) cout << a[c][i] << " ";
    cout << endl;
  }
}
vector<vector<long long> > mul(vector<vector<long long> > a,
                               vector<vector<long long> > b) {
  long long n = a.size();
  long long m = b[0].size();
  vector<vector<long long> > answ(n, vector<long long>(m));
  for (long long c = 0; c < n; c++) {
    for (long long i = 0; i < m; i++) {
      long long kol = 0;
      for (long long j = 0; j < b.size(); j++) {
        kol += a[c][j] * b[j][i];
        kol %= inf;
      }
      answ[c][i] = kol;
    }
  }
  return answ;
}
vector<vector<long long> > binpow(vector<vector<long long> > &a, long long n) {
  if (n == 1) return a;
  if (n % 2) {
    return mul(binpow(a, n - 1), a);
  } else {
    dop = binpow(a, n / 2);
    return mul(dop, dop);
  }
}
int main() {
  long long kol_vo, k;
  cin >> kol_vo >> k;
  vector<vector<long long> > v;
  vector<long long> h(3);
  for (long long c = 0; c < kol_vo; c++) {
    long long a, l, r;
    cin >> a >> l >> r;
    h[0] = l;
    h[1] = 0;
    h[2] = a;
    v.push_back(h);
    h[0] = r;
    h[1] = 1;
    h[2] = a;
    v.push_back(h);
  }
  sort(v.begin(), v.end());
  vector<vector<long long> > all(3, vector<long long>(3)), now;
  all[0][0] = 1;
  all[0][1] = 1;
  all[0][2] = 0;
  all[1][0] = 1;
  all[1][1] = 1;
  all[1][2] = 1;
  all[2][0] = 0;
  all[2][1] = 1;
  all[2][2] = 1;
  now = all;
  long long last = 1;
  vector<vector<long long> > answ(1, vector<long long>(3, 0));
  answ[0][1] = 1;
  bool b = false;
  vector<int> d(3);
  for (long long c = 0; c < v.size(); c++) {
    if (v[c][0] != last) b = false;
    long long kol = v[c][0] - last - 1;
    if (v[c][1] == 1) {
      answ = mul(answ, now);
      b = true;
    }
    if (kol > 0) {
      vector<vector<long long> > bm = binpow(now, kol);
      answ = mul(answ, bm);
    }
    if (v[c][1] == 0) {
      for (long long i = 0; i < 3; i++) now[i][v[c][2] - 1] = 0;
      d[v[c][2] - 1]++;
    } else {
      d[v[c][2] - 1]--;
      if (d[v[c][2] - 1] == 0) {
        for (long long i = 0; i < 3; i++)
          now[i][v[c][2] - 1] = all[i][v[c][2] - 1];
      }
    }
    if (!b && v[c + 1][0] != v[c][0]) {
      answ = mul(answ, now);
    }
    last = v[c][0];
  }
  long long kol = k - last;
  vector<vector<long long> > bm = binpow(now, kol);
  answ = mul(answ, bm);
  cout << answ[0][1];
  return 0;
}
