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>
using namespace std;
#define fo(i,a,b) for(int i=a;i<=b;i++)
typedef long long ll;

const int mod=1e9+7,N=308;
struct node{int to,nxt,l;}c[N];
int head[N],tot;
int n,m,l[N],a[N],f[N][N][N],ans;

void add(int x,int y,int z)
{
	c[++tot].to=y;	c[tot].l=z;
	c[tot].nxt=head[x];	head[x]=tot;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int l,r,x;
		scanf("%d%d%d",&l,&r,&x);
		if(r==1&&x!=1)
		{
			puts("0");
			return 0;
		}
		add(r,l,x);
	}
	f[1][0][0]=1;
	for(int i=1;i<n;i++)
	{
		for(int j=0;j<=i;j++)
			for(int k=j;k<=i;k++)
				if(f[i][j][k])
				{
					if(j||k) (f[i+1][k][i]+=f[i][j][k])%=mod,
								 (f[i+1][j][i]+=f[i][j][k])%=mod;
					else (f[i+1][0][i]+=f[i][j][k])%=mod;
					(f[i+1][j][k]+=f[i][j][k])%=mod;
				}
		for(int t=head[i+1];t;t=c[t].nxt)
		{
			int l=c[t].to;
			for(int j=0;j<=i;j++)
				for(int k=j;k<=i;k++)
				{
					int r=1+(j>=l)+(k>=l);
					if(r!=c[t].l) f[i+1][j][k]=0;
				}
		}
	}
	for(int j=0;j<=n;j++)
		for(int k=j;k<=n;k++)
		{
			int tmp=0;
			if(j) tmp++;	if(k) tmp++;
			if(!tmp) (ans+=(ll)f[n][j][k]*3%mod)%=mod;
			else (ans+=(ll)f[n][j][k]*6%mod)%=mod;
		}
	(ans+=mod)%=mod;
	printf("%d\n",ans);
	return 0;
}