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 N = 1e4 + 100, MOD = 1e9 + 7;
pair<long long, long long> events[2 * N];
array<array<long long, 3>, 3> mult(array<array<long long, 3>, 3> x,
                                   array<array<long long, 3>, 3> y) {
  array<array<long long, 3>, 3> res{array<long long, 3>{0, 0, 0},
                                    array<long long, 3>{0, 0, 0},
                                    array<long long, 3>{0, 0, 0}};
  for (long long i = 0; i < 3; i++)
    for (long long j = 0; j < 3; j++)
      for (long long k = 0; k < 3; k++)
        res[i][k] = (res[i][k] + x[i][j] * y[j][k]) % MOD;
  return res;
}
array<array<long long, 3>, 3> ppow(array<array<long long, 3>, 3> x,
                                   long long y) {
  array<array<long long, 3>, 3> res{array<long long, 3>{1, 0, 0},
                                    array<long long, 3>{0, 1, 0},
                                    array<long long, 3>{0, 0, 1}};
  while (y) {
    if (y & 1) res = mult(res, x);
    y >>= 1;
    x = mult(x, x);
  }
  return res;
}
void print(array<array<long long, 3>, 3> x) {
  for (long long i = 0; i < 3; i++, cout << "\n")
    for (long long j = 0; j < 3; j++) cout << x[i][j] << " ";
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  events[0] = {2, 0};
  for (long long i = 1; i <= n; i++) {
    long long a, l, r;
    cin >> a >> l >> r;
    events[2 * i - 1] = {l, a};
    events[2 * i] = {r + 1, -a};
  }
  events[2 * n + 1] = {m + 1, 0};
  long long cnt[4] = {};
  array<array<long long, 3>, 3> res{array<long long, 3>{1, 0, 0},
                                    array<long long, 3>{0, 1, 0},
                                    array<long long, 3>{0, 0, 1}};
  sort(events, events + 2 * n + 2);
  for (long long i = 0; i <= 2 * n; i++) {
    if (events[i].second > 0)
      cnt[events[i].second]++;
    else
      cnt[-events[i].second]--;
    array<array<long long, 3>, 3> cur{array<long long, 3>{0, 0, 0},
                                      array<long long, 3>{0, 0, 0},
                                      array<long long, 3>{0, 0, 0}};
    if (!cnt[1]) cur[0][0] = cur[1][0] = 1;
    if (!cnt[2]) cur[0][1] = cur[1][1] = cur[2][1] = 1;
    if (!cnt[3]) cur[1][2] = cur[2][2] = 1;
    res = mult(res, ppow(cur, events[i + 1].first - events[i].first));
  }
  cout << res[1][1];
}
