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 N 500
#define M 200000

int n , m ;
int dp[N][N] , a[M] , b[M] , vis[M] ;

int main()
{
    scanf("%d%d" , &n , &m ) ;
    for(int i = 1 ; i <= m ; i++ ){
        scanf("%d%d" , &a[i] , &b[i] ) ;
    }
    for(int i = 1 ; i <= n ; i++ ){
        dp[i][i] = 1 ;
        for(int j = m ; j >= 1 ; j-- ){
            if( dp[i][a[j]] && dp[i][b[j]] ) vis[i] = 1 ;
            else if( dp[i][a[j]] ) dp[i][b[j]] = 1 ;
            else if( dp[i][b[j]] ) dp[i][a[j]] = 1 ;
        }
    }
    int ans = 0 ;
    for(int i = 1 ; i <= n ; i++ ){
        if( vis[i] ) continue ;
        for(int j = i + 1 ; j <= n ; j++ ){
            if( vis[j] ) continue ;
            int t = 0 ;
            for(int k = 1 ; k <= n ; k++ ){
                if( dp[i][k] && dp[j][k] ) t = 1 ;
            }
            ans += ( t == 0 ) ;
        }
    }
    printf("%d\n" , ans ) ;
    return 0 ;
}