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 next Next
#define last Last
#define int long long
const int mod=1e9+7;
const int N=305;
int n,m,f[N][N][N];
struct node{
	int x,y;
};
vector<node>g[N];
/*char buf[1<<21],*p1=buf,*p2=buf;
inline int gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}*/
#define gc getchar
inline int read()
{
	int ret=0,f=0;char c=gc();
	while(!isdigit(c)){if(c=='-')f=1;c=gc();}
	while(isdigit(c)){ret=ret*10+c-48;c=gc();}
	if(f)return -ret;return ret;
}
signed main()
{
	n=read();m=read();
	for(int i=1;i<=m;i++)
	{
		int l=read(),r=read(),x=read();
		g[r].push_back((node){l,x});
	}
	f[0][0][0]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<i;j++)
			for(int k=0;k<=j;k++)
			{
				f[i][j][k]=(f[i][j][k]+f[i-1][j][k])%mod;
				f[i][i-1][k]=(f[i][i-1][k]+f[i-1][j][k])%mod;
				f[i][i-1][j]=(f[i][i-1][j]+f[i-1][j][k])%mod;
				for(int p=0;p<g[i].size();p++)
				{
					node u=g[i][p];
					int gs=1+(j>=u.x)+(k>=u.x);
					if(gs!=u.y)f[i][j][k]=0;
				}
			}
	int ans=0;
	for(int i=0;i<n;i++)
		for(int j=0;j<=i;j++)ans=(ans+f[n][i][j])%mod;
	cout<<ans;
	return 0;
}