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;
using ll = long long;

int N, M;

vector<int> end1[305], end2[305], end3[305];

ll dp[305][305][305];
bool seen[305][305][305]; 

ll f(int n, int a, int b) {
    if (n == N+1) return 1;
    if (seen[n][a][b]) return dp[n][a][b];
    seen[n][a][b] = 1;
    bool agood = 1, bgood = 1, cgood = 1;
    for (int s : end1[n]) {
        if (s <= b) return 0;
        if (s < n) agood = bgood = 0;
    }
    for (int s : end2[n]) {
        if (s <= a || s == n) return 0;
        if (s <= b) agood = 0;
        else if (s < n) cgood = 0;
    }
     for (int s : end3[n]) {
         if (s > b) return 0;
         if (s > a) bgood = cgood = 0;
     }
     if (agood) dp[n][a][b] += f(n+1,b,n-1);
     if (bgood) dp[n][a][b] += f(n+1,a,n-1);
     if (cgood) dp[n][a][b] += f(n+1,a,b);
     dp[n][a][b] %= 1000000007;
     return dp[n][a][b];
}

int main() {
    scanf("%d %d", &N, &M);
    for (int i = 0, l, r, x; i < M; i++) {
        scanf("%d %d %d", &l, &r, &x);
        if (x == 1) end1[r].push_back(l);
        if (x == 2) end2[r].push_back(l);
        if (x == 3) end3[r].push_back(l);
    }
    printf("%lld\n", f(1,0,0));
}