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 p = 1000000007;
struct mat {
  long long val[3][3];
  mat operator*(const mat& rhs) const {
    mat res;
    memset(res.val, 0, sizeof res.val);
    for (int i = 0; i < (3); i++)
      for (int j = 0; j < (3); j++)
        for (int k = 0; k < (3); k++) {
          res.val[i][k] += val[i][j] * rhs.val[j][k];
          res.val[i][k] %= p;
        }
    return res;
  }
  void print() {
    for (int i = 0; i < (3); i++) {
      for (int j = 0; j < (3); j++) cout << val[i][j] << ' ';
      cout << endl;
    }
  }
};
mat trans = {{{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}};
mat powmod(mat b, long long e) {
  mat r = {{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}};
  while (e) {
    if (e & 1) r = r * b;
    b = b * b;
    e >>= 1;
  }
  return r;
}
int n;
long long m;
int blk[3];
vector<tuple<long long, int, int>> events;
int main() {
  cin >> n >> m;
  while (n--) {
    int a;
    long long l, r;
    cin >> a >> l >> r;
    events.push_back(make_tuple(l - 1, a - 1, 1));
    events.push_back(make_tuple(r, a - 1, -1));
  }
  events.push_back(make_tuple(m, 0, 0));
  sort((events).begin(), (events).end());
  long long lpos = 1;
  mat cur = {{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}};
  for (auto& e : events) {
    long long pos;
    int a, delta;
    tie(pos, a, delta) = e;
    ((void)0);
    mat t = trans;
    for (int i = 0; i < (3); i++)
      if (blk[i])
        for (int j = 0; j < (3); j++) t.val[i][j] = 0;
    cur = powmod(t, pos - lpos) * cur;
    lpos = pos;
    blk[a] += delta;
  }
  cout << cur.val[1][1] << endl;
  return 0;
}
