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>
#define mo 1000000007
int max(int a,int b){return a>b?a:b;}
bool cant[305][305][305];int n,m,f[305][305][305],ans;
void add(int&a,int b){a+=b;if(a>=mo)a-=mo;}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1,l,r,t;i<=m;i++)
	{
		scanf("%d%d%d",&l,&r,&t);
		for(int j=l+1;j<=r;j++)
		 if(t==1)
		 {
		 	for(int y=l;y<j;y++)
		 	 for(int x=0;x<y;x++)
			  cant[x][y][j]=1; 
		 }else if(t==2)
		 {
		 	for(int y=l+1;y<j;y++)
		 	 for(int x=l;x<y;x++)
		 	  cant[x][y][j]=1;
		 }
		if(t==2)
		{
			for(int y=0;y<l;y++)
			 for(int x=0;x<=max(y-1,0);x++)
			  cant[x][y][r]=1;
		}else
		if(t==3)
		{
			for(int y=0;y<l;y++)
			 for(int x=0;x<=max(y-1,0);x++)
			  cant[x][y][r]=1;
			for(int y=l;y<r;y++)
			 for(int x=0;x<l;x++)
			  cant[x][y][r]=1;
		}
	}f[0][0][0]=1;
	for(int k=0;k<n;k++)
	 for(int j=0;j<=max(k-1,0);j++)
	  for(int i=0;i<=max(j-1,0);i++)
	   if(!cant[i][j][k]&&f[i][j][k])
	    add(f[i][j][k+1],f[i][j][k]),
	    add(f[j][k][k+1],f[i][j][k]),
	    add(f[i][k][k+1],f[i][j][k]);
	for(int j=0;j<=max(n-1,0);j++)
	 for(int i=0;i<=max(j-1,0);i++)
	  if(!cant[i][j][n]&&f[i][j][n])add(ans,f[i][j][n]);
	printf("%d\n",ans);
}