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;
template <typename T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
const long long MOD = 1e9 + 7;
using mat = vector<vector<long long>>;
inline mat mul(const mat& a, const mat& b) {
  int A = a.size(), B = b.size(), C = b[0].size();
  mat res(A, vector<long long>(C));
  for (int i = 0; i < A; i++) {
    for (int j = 0; j < C; j++) {
      for (int k = 0; k < B; k++) {
        res[i][j] = (res[i][j] + a[i][k] * b[k][j]) % MOD;
      }
    }
  }
  return res;
}
inline mat pow(mat a, long long b) {
  int A = a.size();
  mat res(A, vector<long long>(A));
  for (int i = 0; i < A; i++) res[i][i] = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int N;
long long M;
vector<pair<pair<long long, int>, int>> ev;
const mat tr2{{1, 1}, {1, 1}};
const mat tr3{{1, 1, 0}, {1, 1, 1}, {0, 1, 1}};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 0; i < N; i++) {
    int a;
    long long l, r;
    cin >> a >> l >> r;
    ev.push_back({{l, a - 1}, 1});
    ev.push_back({{r + 1, a - 1}, -1});
  }
  ev.push_back({{M + 1, 0}, 0});
  sort(ev.begin(), ev.end());
  int cnt[3]{};
  long long res[3]{0, 1, 0};
  long long last = 2;
  auto it = ev.begin();
  while (it != ev.end()) {
    long long r = it->first.first;
    if (last == 2 && r == 2) {
    } else {
      if (cnt[0] == 0 && cnt[1] == 0 && cnt[2] == 0) {
        mat tmp = mul(pow(tr3, r - last), {{res[0]}, {res[1]}, {res[2]}});
        res[0] = tmp[0][0], res[1] = tmp[1][0], res[2] = tmp[2][0];
      } else if (cnt[0] == 0 && cnt[1] == 0) {
        mat tmp = mul(
            pow(tr2, r - last - 1),
            {{(res[0] + res[1]) % MOD}, {(res[0] + res[1] + res[2]) % MOD}});
        res[0] = tmp[0][0], res[1] = tmp[1][0], res[2] = 0;
      } else if (cnt[1] == 0 && cnt[2] == 0) {
        mat tmp = mul(
            pow(tr2, r - last - 1),
            {{(res[0] + res[1] + res[2]) % MOD}, {(res[1] + res[2]) % MOD}});
        res[1] = tmp[0][0], res[2] = tmp[1][0], res[0] = 0;
      } else if (cnt[0] == 0 && cnt[2] == 0) {
        res[0] = (res[0] + res[1]) % MOD;
        res[2] = (res[2] + res[1]) % MOD;
        res[1] = 0;
      } else if (cnt[0] == 0) {
        res[0] = (res[0] + res[1]) % MOD, res[1] = res[2] = 0;
      } else if (cnt[1] == 0) {
        res[1] = (res[0] + res[1] + res[2]) % MOD, res[0] = res[2] = 0;
      } else if (cnt[2] == 0) {
        res[2] = (res[2] + res[1]) % MOD, res[0] = res[1] = 0;
      } else {
        res[0] = res[1] = res[2] = 0;
      }
    }
    if (res[0] == 0 && res[1] == 0 && res[2] == 0) break;
    while (it != ev.end() && it->first.first == r) {
      cnt[it->first.second] += it->second;
      it++;
    }
    last = r;
  }
  cout << res[1] << '\n';
  return 0;
}
