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;
template <typename T>
void gmin(T &a, T b) {
  a = a < b ? a : b;
}
template <typename T>
void gmax(T &a, T b) {
  a = a > b ? a : b;
}
const int p = 1000000007;
struct M {
  long long t[3][3];
  inline long long *operator[](int x) { return t[x]; }
  M() { memset(t, 0, sizeof(t)); }
};
M operator*(M a, M b) {
  M c;
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
      for (int k = 0; k < 3; k++) c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % p;
  return c;
}
M pow(M a, long long x) {
  M c;
  c[0][0] = c[1][1] = c[2][2] = 1;
  for (; x; x >>= 1, a = a * a)
    if (x & 1) c = c * a;
  return c;
}
const int MAXN = 20010;
pair<pair<long long, int>, int> t[MAXN];
int tot, cnt[3];
int n;
long long m;
M ans;
int main() {
  cin >> n >> m;
  t[tot++] = make_pair(make_pair(1, 0), 1);
  for (int i = 0; i < n; i++) {
    int a;
    long long l, r;
    cin >> a >> l >> r;
    a--;
    t[tot++] = make_pair(make_pair(l, -1), a);
    t[tot++] = make_pair(make_pair(r + 1, 1), a);
  }
  sort(t, t + tot);
  ans[0][0] = ans[0][2] = 1;
  ans[0][1] = p - 1;
  M c;
  c[0][0] = c[0][1] = c[1][0] = c[1][1] = c[1][2] = c[2][1] = c[2][2] = 1;
  for (int i = 1; i < tot; i++) {
    int op = t[i].first.second, a = t[i].second;
    ans = ans * pow(c, t[i].first.first - t[i - 1].first.first);
    cnt[a] -= op;
    for (int k = 0; k < 3; k++) c[k][a] = cnt[a] <= 0 && (abs(k - a) <= 1);
  }
  ans = ans * pow(c, m - t[tot - 1].first.first + 1);
  cout << ans[0][1];
  return 0;
}
