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;
struct mat {
  long long x[3][3];
} mi, cur, ba;
mat operator*(const mat &a, const mat &b) {
  mat re;
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++) {
      re.x[i][j] = 0;
      for (int k = 0; k < 3; k++)
        re.x[i][j] = (re.x[i][j] + a.x[i][k] * b.x[k][j]) % 1000000007;
    }
  return re;
}
mat po(mat b, long long p) {
  mat re = mi;
  while (p) {
    if (p % 2) re = re * b;
    b = b * b;
    p /= 2;
  }
  return re;
}
int n, cnt[4], ta;
long long m, la = 2, tb, tc;
vector<pair<long long, int> > v;
int main() {
  for (int i = 0; i < 3; i++) mi.x[i][i] = 1;
  cur = mi;
  scanf("%d%lld", &n, &m);
  while (n--) {
    scanf("%d%lld%lld", &ta, &tb, &tc);
    v.push_back(make_pair(tb, ta));
    v.push_back(make_pair(tc + 1, -ta));
  }
  v.push_back(make_pair(m + 1, 0));
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    for (int j = 0; j < 3; j++)
      for (int k = 0; k < 3; k++)
        ba.x[j][k] = (abs(j - k) <= 1 && cnt[j + 1] == 0);
    cur = po(ba, v[i].first - la) * cur;
    la = v[i].first;
    if (v[i].second > 0)
      cnt[v[i].second]++;
    else if (v[i].second < 0)
      cnt[-v[i].second]--;
  }
  printf("%lld\n", cur.x[1][1]);
  return 0;
}
