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<cstring>
#include<vector>
using namespace std;
const int _=305;
const int mod=1e9+7;
inline int read()
{
	char ch='!';int z=1,num=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')z=-1,ch=getchar();
	while(ch<='9'&&ch>='0')num=(num<<3)+(num<<1)+ch-'0',ch=getchar();
	return z*num;
}
int n,m,f[_][_][_];
struct ha{int l,x;};
vector<ha>q[_];
bool check(int p,int p1,int p2)
{
	int len=q[p].size();
	for(int i=0;i<len;++i)
	{
		ha u=q[p][i];int x=1;
		if(p1>=u.l)x++;
		if(p2>=u.l)x++;
		if(x!=u.x)return 0;
	}
	return 1;
}
void add(int &a,int b){a+=b;if(a>=mod)a-=mod;}
int main()
{
	n=read();m=read();
	for(int i=1;i<=m;++i)
	{
		int l=read(),r=read(),x=read();
		q[r].push_back((ha){l,x});
	}
	f[0][0][0]=1;
	for(int i=0;i<n;++i)
		for(int j=0;j<i||!j;++j)
			for(int k=0;k<j||!k;++k)
			{
				if(!f[i][j][k])continue;
				if(check(i+1,j,k))add(f[i+1][j][k],f[i][j][k]);
				if(check(i+1,i,j))add(f[i+1][i][j],f[i][j][k]);
				if(check(i+1,i,k))add(f[i+1][i][k],f[i][j][k]);
			}
	int ans=0;
	for(int j=0;j<n;++j)
		for(int k=0;k<j||!k;++k)
			add(ans,f[n][j][k]);
	printf("%d\n",ans);
	return 0;
}