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 = 200010;
const long long mod = 1e9 + 7;
vector<long long> num;
struct mat {
  long long a[4][4]{};
  void f() {
    for (long long i = 1; i <= 3; i++) a[i][i] = 1;
  }
  void init() {
    a[1][1] = a[1][2] = a[2][1] = a[2][2] = a[2][3] = a[3][2] = a[3][3] = 1;
  }
  friend mat operator*(mat c1, mat c2) {
    mat c3;
    for (long long i = 1; i <= 3; i++)
      for (long long j = 1; j <= 3; j++)
        for (long long k = 1; k <= 3; k++) {
          c3.a[i][j] = (c3.a[i][j] + c1.a[i][k] * c2.a[k][j]) % mod;
        }
    return c3;
  }
  friend mat operator^(mat c1, long long f1) {
    mat c2;
    c2.f();
    for (; f1; f1 >>= 1, c1 = c1 * c1)
      if (f1 & 1) c2 = c2 * c1;
    return c2;
  }
};
long long find(long long x) {
  return lower_bound(num.begin(), num.end(), x) - num.begin();
}
struct T {
  long long pos, l, r;
} b[N];
long long c[4][N];
signed main() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    long long pos, l, r;
    cin >> pos >> l >> r;
    b[i] = {pos, l, r};
    num.push_back(l - 1);
    num.push_back(r);
  }
  num.push_back(1);
  num.push_back(m);
  sort(num.begin(), num.end());
  num.erase(unique(num.begin(), num.end()), num.end());
  for (long long i = 1; i <= n; i++) {
    long long L = find(b[i].l);
    long long R = find(b[i].r);
    c[b[i].pos][L]++;
    c[b[i].pos][R + 1]--;
  }
  long long st[4]{};
  mat ans;
  ans.a[2][1] = 1;
  for (long long i = 1; i < num.size(); i++) {
    mat t;
    t.init();
    for (long long j = 1; j <= 3; j++) {
      st[j] += c[j][i];
      if (st[j]) t.a[j][1] = t.a[j][2] = t.a[j][3] = 0;
    }
    t = t ^ (num[i] - num[i - 1]);
    ans = t * ans;
  }
  cout << ans.a[2][1] << '\n';
}
