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<cstring>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<vector>
#include<iostream>
#include<set>
#include<map>
#include<functional>
#include<queue>
#include<string>

using namespace std;

const int MX=311,md=1000000007;
typedef pair<int,int> pii;
#define mp make_pair
#define se second
#define fr first
vector<pii>G[MX];
typedef vector<pii>::iterator vit;
int dp[MX][MX][MX];
int n,m;

int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int l,r,x;scanf("%d%d%d",&l,&r,&x);
		G[r].push_back(mp(l,x));
	}
	dp[0][0][0]=1;
	int ans=0;
	for(int i=1;i<=n;i++){
		for(int j=0;j<i;j++)
			for(int k=0;k<i;k++){
				bool flg=true;
				for(vit it=G[i].begin();it!=G[i].end();it++){
					if((i>=it->fr)+(j>=it->fr)+(k>=it->fr)!=it->se){flg=false;break;}
				}
				if(!flg)continue;
				if(j==i-1){
					for(int l=0;l<=i-1;l++)dp[i][j][k]=(dp[i][j][k]+dp[j][l][k])%md;
				}else if(k==i-1){
					for(int l=0;l<=i-1;l++)dp[i][j][k]=(dp[i][j][k]+dp[k][j][l])%md;
				}else{
					dp[i][j][k]=dp[i-1][j][k];
				}
				if(i==n)ans=(ans+dp[i][j][k])%md;
			}
	}
	printf("%d\n",((ans+ans)%md+ans)%md);
	return 0;
}