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;
using ll=long long;
static const ll mod=1000000007;
ll N,M;
ll dp[305][305][305];
int main(){
  cin>>N>>M;
  vector<vector<pair<ll,ll>>>X(N+1);bool y=true;
  for(int i=0;i<M;i++){
  ll l,r,x;cin>>l>>r>>x;
    if(l==r && x!=1)y=false;
    pair<ll,ll>p(l,x);
    X[r].push_back(p);
  }dp[0][0][0]=1;
  if(!y)
    cout<<0<<endl;
  else{
  for(int i=0;i<N;i++){
  for(int j=0;j<=i;j++)
  for(int k=0;k<=j;k++){
  dp[i+1][j][k]=(dp[i+1][j][k]+dp[i][j][k])%mod;
  dp[i+1][i][k]=(dp[i+1][i][k]+dp[i][j][k])%mod;
  dp[i+1][i][j]=(dp[i+1][i][j]+dp[i][j][k])%mod;
  
  
  }for(auto p:X[i+1]){
  ll l=p.first;
    ll x=p.second;
    if(x==1){
    for(int j=0;j<=i+1;j++)
      for(int k=0;k<=j;k++)
        if(l<=j)dp[i+1][j][k]=0;
    }else if(x==2){
    for(int j=0;j<=i+1;j++)
      for(int k=0;k<=j;k++)
        if(l<=k || j<l)dp[i+1][j][k]=0;
    }else{
     for(int j=0;j<=i+1;j++)
      for(int k=0;k<=j;k++)
        if(k<l)dp[i+1][j][k]=0;
    }
  }
    
    
  }
  ll ans=0;
  for(int j=0;j<=N;j++)
    for(int k=0;k<=j;k++)
      ans=(ans+dp[N][j][k])%mod;
  cout<<ans<<endl;
  }
  return 0;
}