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>
#define mod 1000000007
using namespace std;
int n,m,ans;
int dp[305][305][305];
struct query{
	int l,r,x;
}e[305];

int cmp(query u,query v){
	return u.r<v.r;
}

int main(){
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%d %d %d",&e[i].l,&e[i].r,&e[i].x);	
	}
	sort(e+1,e+m+1,cmp);
	dp[1][0][0]=3;
	int R=0;
	for(int i=1;i<=n;i++){
		while(e[R+1].r==i&&R<m){
			R++;
			for(int j=0;j<=i-1;j++){
				for(int k=j;k<=i-1;k++){
					if(e[R].x==1&&(j>=e[R].l||k>=e[R].l)) dp[i][j][k]=0;
					if(e[R].x==2&&((j>=e[R].l&&k>=e[R].l)||(j<e[R].l&&k<e[R].l))) dp[i][j][k]=0;
					if(e[R].x==3&&(j<e[R].l||k<e[R].l)) dp[i][j][k]=0;
				}
			}
		}
		if(i==n){
			for(int j=0;j<=i-1;j++){
				for(int k=j;k<=i-1;k++){
					ans+=dp[i][j][k];if(ans>=mod) ans-=mod;
				}
			}
			printf("%d\n",ans);return 0;
		}
		for(int j=0;j<=i-1;j++){
			for(int k=j;k<=i-1;k++){
				dp[i+1][j][k]+=dp[i][j][k];if(dp[i+1][j][k]>=mod) dp[i+1][j][k]-=mod;
				dp[i+1][k][i]+=dp[i][j][k];if(dp[i+1][k][i]>=mod) dp[i+1][k][i]-=mod;
				dp[i+1][j][i]+=dp[i][j][k];if(dp[i+1][j][i]>=mod) dp[i+1][j][i]-=mod;
			}
		}
	}
	
	return 0;
}