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;
int n, cnt = 0;
long long m;
int a[200007];
long long l[200007], r[200007];
int g[5][5], ans[5][5], f[5][5];
int vis[5];
pair<long long, pair<int, int> > q[200007];
void multiply1() {
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++) f[i][j] = 0;
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++)
      for (int k = 1; k <= 3; k++)
        f[i][j] = (f[i][j] + 1LL * g[i][k] * g[k][j]) % 1000000007;
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++) g[i][j] = f[i][j];
}
void multiply2() {
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++) f[i][j] = 0;
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++)
      for (int k = 1; k <= 3; k++)
        f[i][j] = (f[i][j] + 1LL * ans[i][k] * g[k][j]) % 1000000007;
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++) ans[i][j] = f[i][j];
}
void deal(long long k) {
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++) g[i][j] = 0;
  g[1][2] = 1, g[1][1] = 1;
  g[2][1] = 1, g[2][2] = 1, g[2][3] = 1;
  g[3][2] = 1, g[3][3] = 1;
  for (int i = 1; i <= 3; i++)
    if (vis[i]) {
      for (int j = 1; j <= 3; j++) g[j][i] = 0;
    }
  while (k > 0) {
    if (k % 2 == 1) multiply2();
    k /= 2;
    multiply1();
  }
}
int main() {
  scanf("%d%lld", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d%lld%lld", &a[i], &l[i], &r[i]);
    q[++cnt] = make_pair(l[i], make_pair(a[i], 1));
    q[++cnt] = make_pair(r[i] + 1, make_pair(a[i], -1));
  }
  sort(q + 1, q + cnt + 1);
  memset(ans, 0, sizeof(ans));
  for (int i = 1; i <= 3; i++) ans[i][i] = 1;
  memset(vis, 0, sizeof(vis));
  long long now = 2;
  for (int i = 1; i <= cnt; i++) {
    if (q[i].first == now) {
      int pos = q[i].second.first;
      if (q[i].second.second == 1)
        --vis[pos];
      else
        ++vis[pos];
    } else {
      deal(q[i].first - now);
      int pos = q[i].second.first;
      if (q[i].second.second == 1)
        --vis[pos];
      else
        ++vis[pos];
      now = q[i].first;
    }
  }
  deal(m - now + 1);
  printf("%d\n", ans[2][2]);
  return 0;
}
