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<vector>
#include<algorithm>
using namespace std;
#define MAX 330
#define MOD 1000000007
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
void add(int &x,int y){x+=y;if(x>=MOD)x-=MOD;}
int n,m,f[MAX][MAX][MAX],ans;
struct Node{int l,x;};
vector<Node> e[MAX];
bool check(int a,int b,int c)
{
	int r=max(a,max(b,c));
	for(int i=0,t=e[r].size();i<t;++i)
	{
		int l=e[r][i].l,x=e[r][i].x,s=0;
		s=(a>=l)+(b>=l)+(c>=l);
		if(s!=x)return false;
	}
	return true;
}
int main()
{
	n=read();m=read();
	for(int i=1;i<=m;++i)
	{
		int l=read(),r=read(),x=read();
		e[r].push_back((Node){l,x});
	}
	f[0][0][0]=1;
	for(int i=0;i<=n;++i)
		for(int j=0;j<=n;++j)
			for(int k=0;k<=n;++k)
			{
				if(!f[i][j][k])continue;
				if((i&&k&&i==k)||(i&&j&&i==j)||(j&&k&&j==k))continue;
				if(!check(i,j,k)){f[i][j][k]=0;continue;}
				int r=max(i,max(j,k));
				add(f[r+1][j][k],f[i][j][k]);
				add(f[i][r+1][k],f[i][j][k]);
				add(f[i][j][r+1],f[i][j][k]);
				if(r==n)add(ans,f[i][j][k]);
			}
	printf("%d\n",ans);
	return 0;
}