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 long long N = 40000, M = 1e9 + 7;
struct matrix {
  long long x, y, a[3][3];
  matrix operator*(matrix b) {
    matrix c = (matrix){x, b.y, {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}};
    for (long long i = 0; i < x; i++)
      for (long long j = 0; j < b.y; j++)
        for (long long k = 0; k < y; k++)
          c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j] % M) % M;
    return c;
  }
  matrix operator^(long long b) {
    matrix c = (matrix){x, y, {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}}, a = *this;
    for (; b; a = a * a, b >>= 1)
      if (b & 1) c = c * a;
    return c;
  }
} f3 = {3, 3, {{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}},
  f23 = {3, 3, {{1, 1, 0}, {1, 1, 0}, {0, 1, 0}}},
  f21 = {3, 3, {{0, 1, 0}, {0, 1, 1}, {0, 1, 1}}},
  k = {3, 3, {{0, 1, 0}, {0, 0, 0}, {0, 0, 0}}};
struct ANS {
  long long p, f, id;
} p[N];
bool cmp(ANS a, ANS b) { return a.f < b.f; }
long long n, m, z[10];
int main(void) {
  scanf("%I64d%I64d", &n, &m);
  for (long long i = 1; i <= n; i++)
    scanf("%I64d%I64d%I64d", &p[(i - 1) * 2].p, &p[(i - 1) * 2].f,
          &p[(i - 1) * 2 + 1].f),
        p[(i - 1) * 2 + 1].f++, p[(i - 1) * 2 + 1].p = p[(i - 1) * 2].p;
  p[n * 2] = (ANS){0, m + 1, n * 2};
  for (long long i = 0; i <= n * 2; i++) p[i].id = i;
  sort(p, p + n * 2 + 1, cmp);
  matrix f = k * (f3 ^ (p[0].f - 2));
  for (long long i = 0; i < n * 2; i++) {
    if (p[i].id % 2 == 0)
      z[p[i].p]++;
    else
      z[p[i].p]--;
    if (p[i].f == p[i + 1].f) continue;
    long long ans = (z[1] == 0) + (z[2] == 0) + (z[3] == 0);
    if (ans == 3) f = f * (f3 ^ (p[i + 1].f - p[i].f));
    if (ans == 2 && z[1] != 0) f = f * (f21 ^ (p[i + 1].f - p[i].f));
    if (ans == 2 && z[3] != 0) f = f * (f23 ^ (p[i + 1].f - p[i].f));
    if (ans == 2 && z[2] != 0)
      f.a[0][0] += f.a[0][1], f.a[0][2] += f.a[0][1], f.a[0][1] = 0;
    if (ans == 1 && z[1] == 0)
      f.a[0][0] += f.a[0][1], f.a[0][1] = f.a[0][2] = 0;
    if (ans == 1 && z[2] == 0)
      f.a[0][1] += f.a[0][0] + f.a[0][2], f.a[0][0] = f.a[0][2] = 0;
    if (ans == 1 && z[3] == 0)
      f.a[0][2] += f.a[0][1], f.a[0][0] = f.a[0][1] = 0;
    if (ans == 0) {
      printf("0\n");
      return 0;
    }
  }
  printf("%I64d\n", f.a[0][1]);
  return 0;
}
