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 SZ = 3, MOD = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
long long int mul(long long int a, long long int b) { return a * b % MOD; }
int tmp[SZ][SZ];
void mul(int a[SZ][SZ], int b[SZ][SZ], int res[SZ][SZ]) {
  fill(tmp[0], tmp[SZ], 0);
  for (int i = 0; i < int(SZ); i++)
    for (int j = 0; j < int(SZ); j++)
      for (int k = 0; k < int(SZ); k++) add(tmp[i][j], mul(a[i][k], b[k][j]));
  for (int i = 0; i < int(SZ); i++)
    for (int j = 0; j < int(SZ); j++) res[i][j] = tmp[i][j];
}
void powmat(int a[SZ][SZ], long long int n, int res[SZ][SZ]) {
  for (int i = 0; i < int(SZ); i++)
    for (int j = 0; j < int(SZ); j++) res[i][j] = (i == j);
  while (n) {
    if (n & 1)
      mul(res, a, res), n--;
    else
      mul(a, a, a), n /= 2;
  }
}
void print(int m[3][3]) {
  for (int i = 0; i < int(3); i++) {
    for (int j = 0; j < int(3); j++) cerr << m[i][j] << ' ';
    cerr << '\n';
  }
  cerr << '\n';
}
int basic[SZ][SZ] = {{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}, obs[3] = {},
    ways[3][3] = {{0, 0, 0}, {1, 0, 0}, {0, 0, 0}};
void avanzar(long long int d) {
  int m[SZ][SZ], pot[SZ][SZ];
  for (int i = 0; i < int(SZ); i++)
    for (int j = 0; j < int(SZ); j++)
      if (obs[i])
        m[i][j] = 0;
      else
        m[i][j] = basic[i][j];
  powmat(m, d, pot);
  mul(pot, ways, ways);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  long long int m;
  cin >> n >> m;
  vector<pair<long long int, int>> events;
  for (int i = 0; i < int(n); i++) {
    int a;
    long long int l, r;
    cin >> a >> l >> r;
    events.emplace_back(l, -a);
    events.emplace_back(r, a);
  }
  sort(events.begin(), events.end());
  long long int px = 1;
  for (int i = 0; i < int(int((events).size())); i++) {
    avanzar(events[i].first - px - 1);
    px = events[i].first;
    while (i < int((events).size()) && events[i].first == px &&
           events[i].second < 0) {
      obs[abs(events[i].second) - 1]++;
      i++;
    }
    avanzar(1);
    while (i < int((events).size()) && events[i].first == px &&
           events[i].second > 0) {
      obs[events[i].second - 1]--;
      i++;
    }
    i--;
  }
  avanzar(m - px);
  cout << ways[1][0] << '\n';
  return 0;
}
