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<algorithm>
using namespace std;
const int mm=1000000007;
struct data{
	int l,x,next;
}g[1111];
int tot,i,j,k,n,m,f[311][311][311],h[311],v[311][311][311],tmp,ans;
void adddata(int l,int r,int x){
	g[++tot].l=l;g[tot].x=x;g[tot].next=h[r];h[r]=tot;
}
int check(int x,int a,int b){
	int i,j,k;
	for (i=h[x];i;i=g[i].next){
		j=g[i].l;k=1;
		if (j<=b) k++;
		if (j<=a) k++;
		if (k!=g[i].x) return 0;
	}
	return 1;
}
void update(int x,int a,int b,int w){
	if (!v[x][a][b]) return;
	f[x][a][b]=(f[x][a][b]+w)%mm;
}
int main(){
	scanf("%d%d",&n,&m);memset(f,0,sizeof(f));memset(v,0,sizeof(v));f[0][0][0]=1;
	for (i=1;i<=m;i++) scanf("%d%d%d",&j,&k,&tmp),adddata(j,k,tmp);
	for (i=1;i<=n;i++)
	 for (j=0;j<i;j++)
	  for (k=((j==0)?0:j+1);k<i;k++) v[i][j][k]=check(i,j,k);
	for (i=0;i<=n;i++)
	 for (j=0;j<=i;j++)
	  for (k=((j==0)?0:j+1);k<=i;k++)
	   if (f[i][j][k]){
	   	update(i+1,j,k,f[i][j][k]);
	   	update(i+1,j,i,f[i][j][k]);
	   	update(i+1,k,i,f[i][j][k]);
	   }
	for (ans=0,i=0;i<n;i++)
	 for (j=((i==0)?0:i+1);j<n;j++)
	  if (v[n][i][j])ans=(ans+f[n][i][j])%mm;
	printf("%d\n",ans);
	return 0;
}