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;
inline int read(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 305
#define mod 1000000007
int f[MN][MN][MN],ans;
vector<int>G[MN][4];
int n,m,l[MN],r[MN],x[MN];
void add(int &p,int val){p+=val;if(p>mod)p-=mod;}
int main(){
	n=read(),m=read();
	register int i,j,k,u;
	for(i=1;i<=m;i++)l[i]=read(),r[i]=read(),x[i]=read(),G[r[i]][x[i]].push_back(l[i]);
	f[0][0][0]=1;
	for(i=1;i<=n;i++){
		for(j=0;j<i;j++)for(k=0;k<i;k++)
        for(u=(j<i-1&&k<i-1)?i-1:0;u<i;u++)
        	add(f[i][k][u],f[j][k][u]),
        	add(f[j][i][u],f[j][k][u]),
        	add(f[j][k][i],f[j][k][u]);
        for(k=1;k<=3;k++)
        for(j=0;j<G[i][k].size();j++){
            int L=G[i][k][j];
            for(int a=0;a<=i;a++)
            for(int b=0;b<=i;b++)
            for(int c=(a<i&&b<i)?i:0;c<=i;c++)
            if((a>=L)+(b>=L)+(c>=L)!=k) f[a][b][c]=0;
        }
    }
    for(i=0;i<=n;i++)for(j=0;j<=n;j++) for(k=(i<n&&j<n)?n:0;k<=n;k++) add(ans,f[i][j][k]);
    printf("%d\n",ans);
    return 0;
}