Problem: There are N squares arranged in a row. The squares are numbered 1, 2, ..., N, from left to right.

Snuke is painting each square in red, green or blue. According to his aesthetic sense, the following M conditions must all be satisfied. The i-th condition is:

* There are exactly x_i different colors among squares l_i, l_i + 1, ..., r_i.



In how many ways can the squares be painted to satisfy all the conditions? Find the count modulo 10^9+7.

Constraints

* 1 ≤ N ≤ 300
* 1 ≤ M ≤ 300
* 1 ≤ l_i ≤ r_i ≤ N
* 1 ≤ x_i ≤ 3

Input

Input is given from Standard Input in the following format:


N M
l_1 r_1 x_1
l_2 r_2 x_2
:
l_M r_M x_M


Output

Print the number of ways to paint the squares to satisfy all the conditions, modulo 10^9+7.

Examples

Input

3 1
1 3 3


Output

6


Input

4 2
1 3 1
2 4 2


Output

6


Input

1 3
1 1 1
1 1 2
1 1 3


Output

0


Input

8 10
2 6 2
5 5 1
3 5 2
4 7 3
4 4 1
2 3 1
7 7 1
1 5 2
1 7 3
3 4 2


Output

108

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9+7;
int N, M;
int dp[303][303][303];
vector<pair<int, int> > lx[303];
int main()
{
  scanf("%d%d", &N, &M);
  for(int i=1; i<=N; ++i) lx[i].clear();
  for(int i=0; i<M; ++i)
  {
    int l, r, x; scanf("%d%d%d", &l, &r, &x);
    lx[r].emplace_back(l, x);
  }
  int ans = 0;
  memset(dp, 0, sizeof dp);
  dp[0][0][0] = 1;
  for(int i=0; i<=N; ++i)
    for(int j=0; j<=N; ++j)
      for(int k=0; k<=N; ++k)
      {
        if(dp[i][j][k] == 0) continue;
        int V = max(i, max(j, k));
        for(auto t: lx[V])
        {
          int l, x; tie(l, x) = t;
          if(l <= i && i <= V) --x;
          if(l <= j && j <= V) --x;
          if(l <= k && k <= V) --x;
          if(x != 0){ dp[i][j][k] = 0; break;}
        }
        if(V==N)
        {
          ans += dp[i][j][k]; ans %= MOD;
        }
        else
        {
          dp[V+1][j][k] += dp[i][j][k]; dp[V+1][j][k] %= MOD;
          dp[i][V+1][k] += dp[i][j][k]; dp[i][V+1][k] %= MOD;
          dp[i][j][V+1] += dp[i][j][k]; dp[i][j][V+1] %= MOD;
        }
      }
  printf("%d\n", ans);
}