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 <iostream>
#include <algorithm>

#define mod 1000000007
using namespace std;

long long dp[302][302][302],l,r,x,ans = 0;
bool a[302][302][302];

int main(){
	int i,j,k,n,m;
	cin >> n >> m;
	for(i=0;i<=n;i++){
		for(j=0;j<=n;j++){
			for(k=0;k<=n;k++){
				a[i][j][k] = true;
			}
		}
	}
	for(i=0;i<m;i++){
		cin >> l >> r >> x;
		for(j=0;j<r;j++){
			for(k=0;k<r;k++){
				if((j!=0 && j==k) || j<k){
					a[r][j][k] = false;
				}else if(x==1){
					if(j>=l || k>=l){
						a[r][j][k] = false;
					}
				}else if(x==2){
					if(!((j>=l)^(k>=l))){
						a[r][j][k] = false;
					}
				}else if(x==3){
					if(j<l || k<l){
						a[r][j][k] = false;
					}
				}
			}
		}
	}
	
	dp[0][0][0] = 1;
	for(i=0;i<=n;i++){
		for(j=0;j<=n;j++){
			for(k=0;k<=n;k++){
				if(!a[i][j][k]){
					dp[i][j][k] = 0;
				}
				if((i>j && j>k) || (j==0 && k==0)){
					dp[i+1][j][k] += dp[i][j][k];
					dp[i+1][j][k] = dp[i+1][j][k]%mod;
					dp[i+1][i][k] += dp[i][j][k];
					dp[i+1][i][k] = dp[i+1][i][k]%mod;
					dp[i+1][i][j] += dp[i][j][k];
					dp[i+1][i][j] = dp[i+1][i][j]%mod;
				}
			}
		}
	}
	
	for(i=0;i<=n;i++){
		for(j=0;j<=n;j++){
			ans += dp[n][i][j];
			ans = ans%mod;
		}
	}
	cout << ans << endl;
}