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 siz = 8;
const int MAXN = 1e4 + 5;
const int MOD = 1e9 + 7;
int n, pre, np;
long long m;
int cnt[siz];
struct Matrix {
  int v[siz][siz];
  int x, y;
  void clear() {
    memset(v, 0, sizeof(v));
    x = y = 0;
  }
  void Mmul(Matrix a, Matrix b) {
    clear();
    x = a.x, y = b.y;
    int c = a.y;
    for (int i = 1; i <= x; ++i) {
      for (int j = 1; j <= y; ++j) {
        for (int k = 1; k <= c; ++k) {
          v[i][j] += (long long)a.v[i][k] * b.v[k][j] % MOD;
          v[i][j] %= MOD;
        }
      }
    }
    return;
  }
  Matrix Mpw(Matrix a, long long b) {
    Matrix x;
    x.clear();
    x.x = x.y = a.x;
    for (int i = 1; i <= x.x; ++i) x.v[i][i] = 1;
    while (b) {
      if (b & 1) x.Mmul(x, a);
      b >>= 1;
      a.Mmul(a, a);
    }
    return x;
  }
  void write() {
    for (int i = 1; i <= x; ++i) {
      for (int j = 1; j <= y; ++j) {
        printf("%d ", v[i][j]);
      }
      puts("");
    }
    puts("");
    return;
  }
} A[siz], B;
struct rpg {
  long long x;
  int h;
  bool kd;
} a[MAXN << 1];
struct lint {
  long long siz;
  int kd;
} line[MAXN << 1];
bool cmp(rpg a, rpg b) { return a.x < b.x; }
void build() {
  for (int i = 0; i < siz; ++i) A[i].x = A[i].y = 3;
  for (int i = 0; i < siz; ++i) {
    for (int j = 1; j <= 2; ++j) {
      for (int k = 1; k <= 2; ++k) {
        A[i].v[j][k] = A[i].v[4 - j][4 - k] = 1;
      }
    }
    for (int j = 0; j < 3; ++j) {
      if (i & (1 << j)) {
        for (int k = 1; k <= 3; ++k) {
          A[i].v[k][j + 1] = 0;
        }
      }
    }
  }
  B.x = 3, B.y = B.v[2][1] = 1;
  return;
}
void init() {
  build();
  scanf("%d%I64d", &n, &m);
  for (int i = 1; i <= n; ++i)
    scanf("%d%lld%lld", &a[i].h, &a[i].x, &a[i + n].x),
        a[i + n].kd = 1, a[i + n].h = a[i].h, ++a[i + n].x;
  sort(a + 1, a + (n << 1) + 1, cmp);
  np = line[1].siz = 1;
  for (int i = 1; i <= n << 1; ++i) {
    if (a[i].kd)
      --cnt[a[i].h];
    else
      ++cnt[a[i].h];
    if (a[i + 1].x != a[i].x) {
      int nw = 0;
      for (int j = 1; j <= 3; ++j)
        if (cnt[j]) nw |= 1 << j - 1;
      if (nw != pre) {
        line[np].siz = a[i].x - line[np].siz;
        pre = nw;
        line[++np] = (lint){a[i].x, pre};
      }
    }
  }
  line[np].siz = m - line[np].siz;
  return;
}
void solve() {
  for (int i = 1; i <= np; ++i)
    B.Mmul(A[line[i].kd].Mpw(A[line[i].kd], line[i].siz), B);
  printf("%d\n", B.v[2][1]);
  return;
}
int main() {
  init();
  solve();
  return 0;
}
