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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int N,M,l,r,x;//,r = 0,g = 0,b = 0;
ll dp[310][310][310],inf = 1e9+7;

bool check(int a,int b,int c,int l,int x){
	int A[3] = {a,b,c};
	sort(A,A+3);
	if(x==1) return A[1]<l;
	if(x==2) return A[0]<l && l<=A[1];
	if(x==3) return l<=A[0];
}

int main(){
	cin >> N >> M;
	vector<vector<P>> v(N+1);
	for(int i=1;i<=M;i++){
		cin >> l >> r >> x;
		v[r].push_back({l,x});
	}
	dp[0][0][0] = 1;
	ll ans = 0;
	for(int r=0;r<=N;r++){
		for(int g=0;g<=N;g++){
			for(int b=0;b<=N;b++){
				int m = max(max(r,g),b);
				if(v[m].size()>0){
					for(int i=0;i<v[m].size();i++){
						if(!check(r,g,b,v[m][i].first,v[m][i].second)) dp[r][g][b] = 0;
					}
				}
				dp[m+1][g][b] += dp[r][g][b];
				if(m+1==g || m+1==b) dp[m+1][g][b] = 0;
				dp[r][m+1][b] += dp[r][g][b];
				if(m+1==r || m+1==b) dp[r][m+1][b] = 0;
				dp[r][g][m+1] += dp[r][g][b];
				if(m+1==r || m+1==g) dp[r][g][m+1] = 0;
				dp[m+1][g][b] %= inf; dp[r][m+1][b] %= inf; dp[r][g][m+1] %= inf;
				if(m==N){
					ans += dp[m][g][b]+dp[r][m][b]+dp[r][g][m];
					ans %= inf;
				}
			}
		}
	}
	cout << ans << endl;
}