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;
long long get() {
  long long x = 0;
  char s = getchar();
  while (s < '0' || s > '9') s = getchar();
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
  return x;
}
const long long mo = 1e9 + 7;
struct Matrix {
  long long n, m, c[4][4];
  Matrix() {}
  Matrix(const long long nn, const long long mm) {
    n = nn, m = mm;
    memset(c, 0, sizeof c);
  }
  long long *operator[](const long long x) { return c[x]; }
  void init() {
    for (long long i = 1; i < 4; ++i)
      for (long long j = 1; j < 4; ++j) c[i][j] = 1;
    c[1][3] = c[3][1] = 0;
  }
  Matrix operator*(const Matrix a) {
    Matrix r = Matrix(n, a.m);
    for (long long i = 1; i <= n; ++i)
      for (long long k = 1; k <= m; ++k)
        if (c[i][k])
          for (long long j = 1; j <= a.m; ++j)
            r[i][j] = (r[i][j] + 1ll * c[i][k] * a.c[k][j] % mo) % mo;
    return r;
  }
};
Matrix pow(Matrix x, long long k) {
  Matrix ans = Matrix(x.n, x.m);
  for (long long i = 1; i <= x.n; ++i) ans[i][i] = 1;
  while (k) {
    if (k & 1) ans = ans * x;
    x = x * x;
    k >>= 1;
  }
  return ans;
}
const long long M = 1e4 + 5;
long long n, m, s[M << 1], c[4][M << 1], tot;
struct node {
  long long a, l, r;
} a[M];
signed main() {
  n = get(), m = get();
  for (long long i = 1; i <= n; ++i) {
    a[i].a = get(), a[i].l = get(), a[i].r = get();
    s[++tot] = a[i].l - 1, s[++tot] = a[i].r;
  }
  s[++tot] = 1, s[++tot] = m;
  sort(s + 1, s + tot + 1);
  tot = unique(s + 1, s + tot + 1) - s - 1;
  for (long long i = 1; i <= n; ++i) {
    long long l = lower_bound(s + 1, s + tot + 1, a[i].l) - s;
    long long r = lower_bound(s + 1, s + tot + 1, a[i].r) - s;
    ++c[a[i].a][l], --c[a[i].a][r + 1];
  }
  Matrix ans = Matrix(3, 3);
  ans[2][1] = 1;
  long long ss[4] = {};
  for (long long i = 2; i <= tot; ++i) {
    long long len = s[i] - s[i - 1];
    Matrix tmp = Matrix(3, 3);
    tmp.init();
    for (long long j = 1; j < 4; ++j) {
      ss[j] += c[j][i];
      if (ss[j]) tmp[j][1] = tmp[j][2] = tmp[j][3] = 0;
    }
    tmp = pow(tmp, len);
    ans = tmp * ans;
  }
  printf("%lld", ans[2][1]);
  return 0;
}
