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<cstdio>
#include<vector>
#define ll long long
using namespace std;
#define pb push_back
const int ha=1e9+7,maxn=305;
inline void ADD(int &x,int y){ x+=y; if(x>=ha) x-=ha;}
int f[maxn][maxn][maxn],n,m,k,L,R,ans;
vector<int> g[maxn][4];

inline void update(int p){
	for(int i=1;i<=3;i++)
	    for(int j=g[p][i].size()-1,l;j>=0;j--){
	    	l=g[p][i][j];
	    	
	    	for(int a=0;a<=p;a++)
	    	    for(int b=0;b<=p;b++)
	    	        for(int c=(a<p&&b<p)?p:0;c<=p;c++)
	    	            if((a>=l)+(b>=l)+(c>=l)!=i) f[a][b][c]=0;
		}
}

inline void dp(){
	f[0][0][0]=1;
	for(int i=1;i<=n;i++){
		for(int j=0;j<i;j++)
		    for(int l=0,now;l<i;l++)
		        for(int u=(j<i-1&&l<i-1)?i-1:0;u<i;u++){
		        	now=f[j][l][u];
		        	ADD(f[i][l][u],now);
		        	ADD(f[j][i][u],now);
		        	ADD(f[j][l][i],now);
				}
		
		update(i);
	}
	
	for(int j=0;j<=n;j++)
	    for(int l=0;l<=n;l++)
	        for(int u=(j<n&&l<n)?n:0;u<=n;u++) ADD(ans,f[j][l][u]);
}

int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%d%d%d",&L,&R,&k);
		g[R][k].pb(L);
	}
	
	dp();
	
	printf("%d\n",ans);
	return 0;
}