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 dim = 3;
const vector<vector<long long>> Id = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
long long MOD = 1e9 + 7;
vector<vector<long long>> mul(vector<vector<long long>> A,
                              vector<vector<long long>> B) {
  assert(A[0].size() == B.size());
  vector<vector<long long>> ans(A.size(), vector<long long>(B[0].size(), 0));
  for (int k = 0; k < B.size(); ++k)
    for (int i = 0; i < A.size(); ++i)
      for (int j = 0; j < B[0].size(); ++j)
        (ans[i][j] += A[i][k] * B[k][j]) %= MOD;
  return ans;
}
vector<vector<long long>> mat_pow(vector<vector<long long>> A, long long k) {
  vector<vector<long long>> ans = Id;
  for (vector<vector<long long>> cur = A; k; k >>= 1, cur = mul(cur, cur))
    if (k & 1) ans = mul(ans, cur);
  return ans;
}
void print_mat(vector<vector<long long>> A) {
  for (int i = 0; i < A.size(); ++i)
    for (int j = 0; j < A[i].size(); ++j)
      cout << A[i][j] << " \n"[j == A[i].size() - 1];
}
pair<long long, long long> A[20001];
vector<vector<long long>> trans = {{1, 1, 0}, {1, 1, 1}, {0, 1, 1}},
                          cur_trans = trans;
vector<vector<long long>> zeros = {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}};
int curob[3] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long l, r, a, n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i)
    cin >> a >> l >> r, A[i] = {l, a}, A[i + n] = {r + 1, -a};
  A[2 * n] = {m, 0};
  sort(A, A + 2 * n + 1);
  vector<vector<long long>> cur = {{0}, {1}, {0}};
  long long pre = 1;
  for (int i = 0; i < 2 * n + 1; ++i) {
    tie(l, a) = A[i];
    cur = mul(mat_pow(cur_trans, l - pre - 1), cur);
    if (a > 0) ++curob[a - 1], cur_trans[a - 1] = zeros[a - 1];
    if (a < 0) {
      --curob[-a - 1];
      if (curob[-a - 1] == 0) cur_trans[-a - 1] = trans[-a - 1];
    }
    pre = l - 1;
  }
  cout << (cur[1][0] + cur[0][0] + cur[2][0]) % MOD << '\n';
}
