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 int MAXN = 2e4 + 7;
const long long mod = 1e9 + 7;
long long n, m;
vector<long long> V;
struct node {
  int row;
  long long l, r;
} p[MAXN];
int now[3], obs[3][MAXN];
struct Matrix {
  long long m[3][3];
  Matrix() { memset(m, 0, sizeof(m)); }
  void gao() {
    for (int i = 0; i <= 2; i++)
      for (int j = 0; j <= 2; j++) m[i][j] = 1;
    m[0][2] = m[2][0] = 0;
  }
  void init() { m[0][0] = m[1][1] = m[2][2] = 1; }
  Matrix operator*(const Matrix &a) const {
    Matrix res;
    for (int i = 0; i <= 2; i++)
      for (int j = 0; j <= 2; j++)
        for (int k = 0; k <= 2; k++)
          res.m[i][j] = (res.m[i][j] + m[i][k] * a.m[k][j] % mod) % mod;
    return res;
  }
  Matrix mpow(Matrix a, long long n) {
    Matrix res;
    res.init();
    while (n) {
      if (n & 1) res = res * a;
      a = a * a;
      n >>= 1;
    }
    return res;
  }
  void Print() {
    for (int i = 0; i <= 2; i++) {
      for (int j = 0; j <= 2; j++) cout << m[i][j];
      puts("");
    }
    puts("");
  }
};
int main() {
  int cnt = 0;
  scanf("%I64d%I64d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d%I64d%I64d", &p[i].row, &p[i].l, &p[i].r);
    p[i].row--;
    p[i].r++;
    V.push_back(p[i].l);
    V.push_back(p[i].r);
  }
  V.push_back(1), V.push_back(m);
  sort(V.begin(), V.end());
  V.erase(unique(V.begin(), V.end()), V.end());
  for (int i = 1; i <= n; i++) {
    int row = p[i].row;
    long long l = p[i].l, r = p[i].r;
    int _l = lower_bound(V.begin(), V.end(), l) - V.begin();
    int _r = lower_bound(V.begin(), V.end(), r) - V.begin();
    obs[row][_l]++;
    obs[row][_r]--;
  }
  Matrix ans;
  ans.m[1][1] = 1;
  for (int i = 1; i <= V.size() - 1; i++) {
    long long len = V[i] - V[i - 1];
    Matrix temp;
    temp.gao();
    for (int j = 0; j <= 2; j++) {
      now[j] += obs[j][i - 1];
      if (now[j]) temp.m[j][0] = temp.m[j][1] = temp.m[j][2] = 0;
    }
    temp = temp.mpow(temp, len);
    ans = ans * temp;
  }
  cout << ans.m[1][1] << endl;
  return 0;
}
