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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod=1e9+7;
struct P{
	int l,r,x;
}a[305];
int n,m,f[305][305][305];
bool cmp(P a,P b){
	return a.r<b.r;
}
int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++) scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].x);
	sort(a+1,a+m+1,cmp);
	f[0][0][0]=1;
	for(int i=0,pos=0,pre=0;i<=n;i++){
		while(pos<m&&a[pos+1].r==i) pos++;
		for(int p=0;p<i;p++){
			for(int j=0;j<=p;j++)if(!j||j<p){
				for(int k=pre+1;k<=pos;k++){
					if(a[k].x==1&&a[k].l<=p){
						f[i][p][j]=0;break;
					}
					if(a[k].x==2&&(a[k].l>p||a[k].l<=j)){
						f[i][p][j]=0;break;
					}
					if(a[k].x==3&&!(a[k].l<=j)){
						f[i][p][j]=0;break;
					}
				}
			}
		}
		if(i==n) continue;
		for(int p=0;p<=i;p++)if(!p||p<i){
			for(int j=0;j<=p;j++)if(!j||j<p){
				f[i+1][p][j]=add(f[i+1][p][j],f[i][p][j]);
				f[i+1][i][j]=add(f[i+1][i][j],f[i][p][j]);
				f[i+1][i][p]=add(f[i+1][i][p],f[i][p][j]);
			}
		}
		pre=pos;
	}
	int ans=0;
	for(int i=0;i<=n;i++){
		for(int p=0;p<=n;p++) ans=add(ans,f[n][i][p]);
	}
	printf("%d",ans);
	return 0;
}