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>
using namespace std;
 
#define mod 1000000007
int n,m,l,r,x,ans=0;
vector<pair<int,int> > v[305];
int f[305][305][305];
 
inline bool check(int r,int g,int b){
    int k=max(r,max(g,b));
    for(int i=0;i<v[k].size();i++){
        int cnt=0;
        int l=v[k][i].first,x=v[k][i].second;
        if(r>=l)cnt++;
        if(g>=l)cnt++;
        if(b>=l)cnt++;
        if(cnt!=x)return false;
    }
    return true;
}
 
inline void add(int &x,const int &y){
    x+=y;if(x>=mod)x-=mod;
}
 
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&l,&r,&x);
        v[r].push_back(make_pair(l,x));
    }
    f[0][0][0]=1;
    register int r,g,b;
    for(r=0;r<=n;r++)
    for(g=0;g<=n;g++)
    for(b=0;b<=n;b++){
        if(f[r][g][b]==0)continue;
        if(!check(r,g,b)){
            f[r][g][b]=0;
            continue;
        }
        int k=max(r,max(g,b));
        add(f[k+1][g][b],f[r][g][b]);
        add(f[r][k+1][b],f[r][g][b]);
        add(f[r][g][k+1],f[r][g][b]);
        if(k==n)add(ans,f[r][g][b]);
    }
    printf("%d",ans);
}