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 ii {
  long long num[4][4];
} fr, now;
long long dp[4], dp2[4], n, m, i, j, x, l, r, ans, mod = 1e9 + 7;
map<long long, map<long long, long long> > mp2;
void inc(long long x, long long y) { mp2[x][y]++; }
void dec(long long x, long long y) { mp2[x][y]--; }
ii mul(ii x, ii y) {
  int i, j, k;
  ii ans;
  memset(ans.num, 0, sizeof(ans.num));
  for (i = 1; i <= 3; i++) {
    for (j = 1; j <= 3; j++) {
      for (k = 1; k <= 3; k++) {
        (ans.num[i][j] += x.num[i][k] * y.num[k][j]) %= mod;
      }
    }
  }
  return ans;
}
ii qp(ii x, long long y) {
  ii z;
  memset(z.num, 0, sizeof(z.num));
  int i;
  for (i = 1; i <= 3; i++) z.num[i][i] = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    y /= 2;
    x = mul(x, x);
  }
  return z;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  memset(fr.num, 0, sizeof(fr.num));
  fr.num[1][1] = fr.num[1][2] = fr.num[2][1] = fr.num[2][2] = fr.num[2][3] =
      fr.num[3][3] = fr.num[3][2] = 1;
  for (i = 1; i <= n; i++) {
    cin >> x >> l >> r;
    r++;
    inc(l, x);
    dec(r, x);
  }
  mp2[m] = mp2[m];
  dp[2] = 1;
  now = fr;
  for (map<long long, map<long long, long long> >::iterator it = mp2.begin();
       it != mp2.end(); it++) {
    if (it != mp2.begin()) {
      map<long long, map<long long, long long> >::iterator pre = it;
      pre--;
      for (i = 1; i <= 3; i++) it->second[i] += pre->second[i];
      now = qp(now, it->first - pre->first);
    } else {
      now = qp(now, it->first - 1);
    }
    memset(dp2, 0, sizeof(dp2));
    for (i = 1; i <= 3; i++) {
      for (j = 1; j <= 3; j++) {
        dp2[j] += dp[i] * now.num[i][j];
        dp2[j] %= mod;
      }
    }
    for (i = 1; i <= 3; i++)
      if (!it->second[i])
        dp[i] = dp2[i];
      else
        dp[i] = 0;
    now = fr;
    for (i = 1; i <= 3; i++) {
      for (j = 1; j <= 3; j++) {
        if (it->second[i]) now.num[i][j] = 0;
      }
    }
  }
  cout << dp[2] << endl;
  return 0;
}
