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 MOD = 1000000007;
struct Matrix {
  long long a[3][3];
  void init() {
    a[0][0] = 1;
    a[0][1] = 1;
    a[0][2] = 0;
    a[1][0] = 1;
    a[1][1] = 1;
    a[1][2] = 1;
    a[2][0] = 0;
    a[2][1] = 1;
    a[2][2] = 1;
  }
  Matrix() { memset(a, 0, sizeof(a)); }
  friend Matrix operator*(const Matrix& x, const Matrix& y) {
    Matrix z;
    for (long long i = 0; i <= 2; i++)
      for (long long j = 0; j <= 2; j++)
        for (long long k = 0; k <= 2; k++)
          z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;
    return z;
  }
} d, ans;
struct Segment {
  long long l, r;
} sg[3][20010];
long long n, dif[3][20010], cnt[3], tot, sum[3][20010];
long long m, num[20010];
template <typename type>
void read(type& x) {
  char c = getchar();
  x = 0;
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
}
Matrix pw(Matrix x, long long y) {
  Matrix res;
  for (long long i = 0; i <= 2; i++) res.a[i][i] = 1;
  while (y) {
    if (y & 1) res = res * x;
    x = x * x;
    y >>= 1;
  }
  return res;
}
void debug(Matrix x) {
  for (long long j = 0; j <= 2; j++) {
    for (long long k = 0; k <= 2; k++) {
      cout << x.a[j][k] << " ";
    }
    cout << endl;
  }
}
signed main() {
  read(n);
  read(m);
  num[1] = 1;
  for (long long i = 1; i <= n; i++) {
    long long a, l, r;
    read(a);
    a--;
    read(l);
    read(r);
    num[i << 1] = l;
    num[(i << 1) + 1] = r + 1;
    sg[a][++cnt[a]] = (Segment){l, r + 1};
  }
  num[(n + 1) << 1] = m;
  sort(num + 1, num + ((n + 1) << 1) + 1);
  tot = unique(num + 1, num + ((n + 1) << 1) + 1) - num - 1;
  for (long long i = 0; i <= 2; i++) {
    for (long long j = 1; j <= cnt[i]; j++) {
      long long l = lower_bound(num + 1, num + tot + 1, sg[i][j].l) - num;
      long long r = lower_bound(num + 1, num + tot + 1, sg[i][j].r) - num;
      dif[i][l]++;
      dif[i][r]--;
    }
  }
  ans.a[0][1] = 1;
  for (long long i = 2; i <= tot; i++) {
    d.init();
    for (long long j = 0; j <= 2; j++) {
      sum[j][i] = sum[j][i - 1] + dif[j][i];
      if (sum[j][i - 1]) {
        for (long long k = 0; k <= 2; k++) d.a[k][j] = 0;
      }
    }
    ans = ans * pw(d, num[i] - num[i - 1] - 1);
    d.init();
    for (long long j = 0; j <= 2; j++) {
      if (sum[j][i]) {
        for (long long k = 0; k <= 2; k++) d.a[k][j] = 0;
      }
    }
    ans = ans * d;
  }
  printf("%d\n", ans.a[0][1]);
}
