There are N turkeys. We number them from 1 through N.

M men will visit here one by one. The i-th man to visit will take the following action:

* If both turkeys x_i and y_i are alive: selects one of them with equal probability, then eats it.
* If either turkey x_i or y_i is alive (but not both): eats the alive one.
* If neither turkey x_i nor y_i is alive: does nothing.



Find the number of pairs (i,\ j) (1 ≤ i < j ≤ N) such that the following condition is held:

* The probability of both turkeys i and j being alive after all the men took actions, is greater than 0.

Constraints

* 2 ≤ N ≤ 400
* 1 ≤ M ≤ 10^5
* 1 ≤ x_i < y_i ≤ N

Input

Input is given from Standard Input in the following format:


N M
x_1 y_1
x_2 y_2
:
x_M y_M


Output

Print the number of pairs (i,\ j) (1 ≤ i < j ≤ N) such that the condition is held.

Examples

Input

3 1
1 2


Output

2


Input

4 3
1 2
3 4
2 3


Output

1


Input

3 2
1 2
1 2


Output

0


Input

10 10
8 9
2 8
4 6
4 9
7 8
2 8
1 8
3 4
3 4
2 7


Output

5
#include <bits/stdc++.h> 
using namespace std;
#define int long long

const int M = 402;
const int N = 100002;
int n, m, X_[N], Y_[N];
bool valid[M], killed[M][M];

signed main(){
 
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin>>m>>n;
    for(int i=1;i<=n;i++)
    	cin>>X_[i]>>Y_[i];

    for(int i=1;i<=m;i++){
    	killed[i][i] = true;
    	valid[i] = true;
    	for(int j=n;j>=1;j--)
    		if(killed[i][X_[j]] && killed[i][Y_[j]]){
    			valid[i] = false;
    			break;
    		}
    		else if(killed[i][X_[j]])
    			killed[i][Y_[j]] = true;
    		else if(killed[i][Y_[j]])
    			killed[i][X_[j]] = true;
    }


    int ans = 0;
    for(int i=1;i<=m;i++)
    	for(int j=i+1;j<=m;j++)
    		if(valid[i] && valid[j]){
    			bool good = true;
    			for(int k=1;k<=m;k++)
    				if(killed[i][k] && killed[j][k])
    					good = false;
    			ans += good;
    		}

    cout<<ans;
    return 0;
}




 

