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 = 5e5 + 14;
const int maxmat = 3, mod = 1e9 + 7;
int n;
long long m;
struct mat {
  int a[maxmat][maxmat];
  mat(bool v = 0) {
    memset(a, 0, sizeof a);
    for (int i = 0; i < maxmat; i++) a[i][i] = v;
  }
  int *operator[](int idx) { return a[idx]; }
};
mat operator*(mat &a, mat &b) {
  mat ans;
  for (int i = 0; i < maxmat; i++)
    for (int k = 0; k < maxmat; k++)
      for (int j = 0; j < maxmat; j++)
        ans[i][j] = (ans[i][j] + (long long)a[i][k] * b[k][j]) % mod;
  return ans;
}
mat operator^(mat a, long long b) {
  mat ans(1);
  for (; b; b >>= 1, a = a * a)
    if (b & 1) ans = ans * a;
  return ans;
}
int block[maxmat];
int dp[maxmat];
map<long long, vector<pair<int, int> > > events;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  while (n--) {
    long long row, l, r;
    cin >> row >> l >> r;
    row--, l--;
    events[l].push_back({row, +1});
    events[r].push_back({row, -1});
  }
  events.insert({1, {}});
  events.insert({m, {}});
  dp[1] = 1;
  for (auto it = events.begin(); it != events.end(); it++) {
    if (it != events.begin()) {
      mat step;
      for (int i = 0; i < maxmat; i++)
        for (int j = max(0, i - 1); j < min(i + 2, maxmat); j++)
          if (!block[j]) step[i][j] = 1;
      step = step ^ (it->first - prev(it)->first);
      int old[maxmat];
      memcpy(old, dp, sizeof dp);
      memset(dp, 0, sizeof dp);
      for (int i = 0; i < maxmat; i++)
        for (int j = 0; j < maxmat; j++)
          dp[i] = (dp[i] + (long long)old[j] * step[j][i]) % mod;
    }
    for (auto [row, v] : it->second) block[row] += v;
  }
  cout << dp[1] << '\n';
}
