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 maxz = 3;
const long long mo = 1e9 + 7;
struct mat {
  typedef long long row[maxz];
  long long a[maxz][maxz];
  void clear(int x) {
    for (int i = 0; i < maxz; ++i) {
      for (int j = 0; j < maxz; ++j) {
        a[i][j] = (i == j) ? x : 0;
      }
    }
  }
  row &operator[](int x) { return a[x]; }
  const row &operator[](int x) const { return a[x]; }
  mat &operator*=(const mat &b) {
    long long t[maxz][maxz];
    memset(t, 0, sizeof t);
    for (int i = 0; i < maxz; ++i) {
      for (int k = 0; k < maxz; ++k) {
        for (int j = 0; j < maxz; ++j) {
          (t[i][j] += a[i][k] * b[k][j]) %= mo;
        }
      }
    }
    memcpy(a, t, sizeof a);
    return *this;
  }
} tr[1 << 3];
mat po(mat a, long long b) {
  mat c;
  c.clear(1);
  while (b) {
    if (b & 1) c *= a;
    a *= a;
    b >>= 1;
  }
  return c;
}
int n;
long long m;
map<long long, int> z;
void hit(long long x) {
  if (x <= 0 || x > m) return;
  int p = z.lower_bound(x)->second;
  z[x] = p;
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  tr[0][0][1] = tr[0][1][0] = tr[0][2][1] = tr[0][1][2] = 1;
  tr[0][0][0] = tr[0][1][1] = tr[0][2][2] = 1;
  for (int i = 1; i < 8; ++i) {
    tr[i] = tr[0];
    for (int j = 0; j < 3; ++j) {
      for (int k = 0; k < 3; ++k) {
        if (i & (1 << k)) {
          tr[i][j][k] = 0;
        }
      }
    }
  }
  cin >> n >> m;
  z[1] = z[m] = 0;
  for (int i = 1; i <= n; ++i) {
    int a;
    long long l, r;
    cin >> a >> l >> r;
    --a;
    a = 1 << a;
    hit(l);
    hit(r);
    hit(l - 1);
    hit(r + 1);
    for (auto &x : z) {
      if (l <= x.first && x.first <= r) x.second |= a;
    }
  }
  mat now;
  now.clear(0);
  now[0][1] = 1;
  pair<long long, int> last;
  for (const auto &x : z) {
    if (x.first == 1) {
      now[0][1] = 1;
    } else {
      mat tmp = po(tr[x.second], x.first - last.first);
      now *= tmp;
    }
    last = x;
  }
  cout << now[0][1] << endl;
  return 0;
}
