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>
#define REP(i,n) for(int i=0;i<(int)(n);i++)
#define ALL(x) (x).begin(),(x).end()
#define LL long long
#define pii pair<int,int>

using namespace std;

int main(){

    int N,M;cin>>N>>M;
    vector<pii> sense[N+1];
    REP(i,M){
        int l,r,x;cin>>l>>r>>x;
        sense[r].push_back({l,x});
    }

    LL dp[N+1][N+1][N+1];
    fill(dp[0][0],dp[N+1][0],0);
    REP(i,N+1)dp[0][0][0]=1;

    LL inf = 1e9+7;

    REP(r,N+1)REP(g,N+1)REP(b,N+1){
                int next = max({r,g,b})+1;
                for(pii p :sense[next-1]){
                    int cnt=0;
                    if(r>=p.first)cnt++;
                    if(g>=p.first)cnt++;
                    if(b>=p.first)cnt++;
                    if(cnt!=p.second)dp[r][g][b]=0;
                }
                if(next<=N){
                    dp[next][g][b]=(dp[next][g][b]+dp[r][g][b])%inf;
                    dp[r][next][b]=(dp[r][next][b]+dp[r][g][b])%inf;
                    dp[r][g][next]=(dp[r][g][next]+dp[r][g][b])%inf;
                }
            }


    LL ans =0;
    REP(i,N)REP(j,N)ans = (ans + dp[N][i][j]+dp[i][N][j]+dp[i][j][N])%inf;

    cout<<ans<<endl;

    return 0;
}