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 ll long long
#define myp pair<int,int>
#define myv vector<myp>
using namespace std;
const int N=310,MOD=1e9+7;
int n,m,can[N];
ll f[N][N][N],ans;
myv v[N];
int main(){
	int t1,t2,t3,nowl,nowx;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%d%d%d",&t1,&t2,&t3);
		v[t2].push_back(myp(t1,t3));
	}
	f[1][0][0]=3;
	for(int i=1;i<=n;i++)
		for(int j=0;j<i;j++){
			memset(can,0,sizeof(can));
			for(int k=0;k<v[i].size();k++){
				nowl=v[i][k].first,nowx=v[i][k].second;
				if(nowx==1&&nowl<=j) can[0]+=1,can[n+1]-=1;
				if(nowx==2){
					if(nowl>j) can[0]+=1,can[n+1]-=1;
					else can[nowl]+=1,can[j]-=1;
				}
				if(nowx==3){
					if(nowl>=j) can[0]+=1,can[n+1]-=1;
					else can[0]+=1,can[nowl]-=1;
				}
			}
			for(int k=1;k<=n;k++) can[k]+=can[k-1];
			for(int k=0;k<=j;k++){
				if(j==k&&j&&k||can[k]) continue;
//				if(!f[i][j][k]) f[i][j][k]=6-3*(!j||!k);
				if(i==n) ans=(ans+f[i][j][k])%MOD;
				f[i+1][j][k]=(f[i+1][j][k]+f[i][j][k])%MOD;
				f[i+1][i][k]=(f[i+1][i][k]+f[i][j][k])%MOD;
				f[i+1][i][j]=(f[i+1][i][j]+f[i][j][k])%MOD;
			}
		}
	cout<<ans<<endl;
	return 0;
}