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;
using INT = long long;
const int NN = 27272;
int a[NN], id[NN];
INT l[NN], r[NN], p[NN];
int ed[7], vst[NN][7];
const int Msize = 7;
const int mod = 1000000007;
struct Matrix {
  int n, mat[Msize][Msize];
  void init(int _n, int a = 0) {
    n = _n;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) mat[i][j] = a;
  }
  void initOne(int _n) {
    n = _n;
    memset(mat, 0, sizeof(mat));
    for (int i = 0; i < n; i++) mat[i][i] = 1;
  }
  Matrix operator*(const Matrix &b) const {
    Matrix ret;
    ret.init(n);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          ret.mat[i][j] =
              (ret.mat[i][j] + ((INT)mat[i][k] * b.mat[k][j]) % mod) % mod;
        }
      }
    return ret;
  }
};
Matrix pow_M(Matrix a, INT n) {
  Matrix ret;
  ret.initOne(a.n);
  for (; n; n >>= 1, a = a * a)
    if (n & 1) ret = ret * a;
  return ret;
}
bool cmp(int u, int v) {
  if (l[u] != l[v]) return l[u] < l[v];
  return r[u] < r[v];
}
int main() {
  int n;
  INT m;
  scanf("%d%I64d", &n, &m);
  int cnt = 0;
  p[++cnt] = 1;
  p[++cnt] = 2;
  for (int i = 1; i <= n; i++) {
    scanf("%d%I64d%I64d", a + i, l + i, r + i);
    p[++cnt] = l[i] - 1;
    p[++cnt] = r[i];
    id[i] = i;
  }
  sort(id + 1, id + n + 1, cmp);
  p[++cnt] = m;
  sort(p + 1, p + cnt + 1);
  cnt = unique(p + 1, p + cnt + 1) - p - 1;
  for (int j = 1; j <= n; j++) {
    int i = id[j];
    int it = lower_bound(p + 1, p + cnt + 1, l[i]) - p;
    int th = upper_bound(p + 1, p + cnt + 1, r[i]) - p;
    for (int j = max(ed[a[i]] + 1, it); j < th; j++) vst[j][a[i]] = 1;
    ed[a[i]] = max(ed[a[i]], th - 1);
  }
  Matrix ans;
  ans.init(3, 0);
  ans.mat[1][1] = 1;
  for (int i = 2; i <= cnt; i++) {
    INT N = p[i] - p[i - 1];
    Matrix P;
    P.init(3, 1);
    P.mat[0][2] = P.mat[2][0] = 0;
    for (int j = 1; j <= 3; j++)
      if (vst[i][j]) {
        for (int k = 0; k < 3; k++) P.mat[k][j - 1] = 0;
      }
    ans = ans * pow_M(P, N);
  }
  cout << ans.mat[1][1] << endl;
  return 0;
}
