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 N 305
#define pi pair<int,int>
#define mk make_pair
#define Mod 1000000007
using namespace std;
int n,m,x,y,z,dp[N][N][N];
vector<pi>G[N];
inline bool check(int x,int y,int z){
	int pos=max(x,max(y,z));
	if (x<y) swap(x,y);if (x<z) swap(x,z);
	if (y<z) swap(y,z);
	for (int i=0;i<(int)G[pos].size();i++){
		int l=G[pos][i].first;
		int limit=G[pos][i].second;
		if (limit==3) if (z<l) return 0;
		if (limit==2){
			if (y<l) return 0;
			if (z>=l) return 0;
		}
		if (limit==1) if (y>=l) return 0;
	}
	return 1;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++){
		scanf("%d%d%d",&x,&y,&z);
		G[y].push_back(mk(x,z));
	}
	dp[0][0][0]=1;
	int ans=0;
	for (int i=0;i<=n;i++){
		for (int j=0;j<=n;j++){
			for (int k=0;k<=n;k++){
				if (!dp[i][j][k]) continue;
				int nxt=max(i,max(j,k))+1;
				if (nxt==n+1){
					ans=(ans+dp[i][j][k])%Mod;
					continue;
				}
				if (check(nxt,j,k)) dp[nxt][j][k]=1ll*(dp[nxt][j][k]+dp[i][j][k])%Mod;
				if (check(i,nxt,k)) dp[i][nxt][k]=1ll*(dp[i][nxt][k]+dp[i][j][k])%Mod;
				if (check(i,j,nxt)) dp[i][j][nxt]=1ll*(dp[i][j][nxt]+dp[i][j][k])%Mod;
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}