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 = 1e9 + 7;

int n,m,i,j,k;
long long dp[305][305][305];
vector <pair<int, int> > q[305];

int main(){
	scanf("%d%d", &n, &m);
	
	for(i = 0; i < m; i++){
		int x,y,z;
		scanf("%d%d%d", &x, &y, &z);
		q[y].push_back({x, z});
	}
	
	dp[0][0][0] = 1LL;
	
	long long res = 0;
	
	for(i = 0; i <= n; i++){
		for(j = 0; j <= i; j++){
			if(i != 0 && j == i) continue;
			for(k = 0; k <= j; k++){
				if(j != 0 && k == j) continue;
				if(dp[i][j][k] == 0) continue;
				
				int nxt = i;
				
				for(int ii = 0; ii < q[nxt].size(); ii++){
					int x = q[nxt][ii].first; int y = q[nxt][ii].second;
					if(y == 1){
						if(j >= x)
						dp[i][j][k] = 0LL;
					}
					else if(y == 2){
						if(j < x || k >= x)
						dp[i][j][k] = 0LL;
					}
					else{
						if(k < x)
						dp[i][j][k] = 0LL;
					}
				}
				
				if(i == n)
				res = (res + dp[i][j][k]) % MOD;
				
				dp[nxt + 1][j][k] = (dp[nxt + 1][j][k] + dp[i][j][k]) % MOD;
				dp[nxt + 1][i][k] = (dp[nxt + 1][i][k] + dp[i][j][k]) % MOD;
				dp[nxt + 1][i][j] = (dp[nxt + 1][i][j] + dp[i][j][k]) % MOD;
			}
		}
	}
	
	printf("%lld\n", res);
}