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>

#define MAXN (410)
#define MAXM (100010)

int n, m;
std::pair<int, int> act[MAXM];
std::bitset<MAXN> s[MAXN];

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= m; ++ i){
		scanf("%d%d", &act[i].first, &act[i].second);
	}
	for(int i = 1; i <= n; ++ i){
		s[i][i] = 1;
		for(int j = m; j >= 1; -- j){
			int x = act[j].first, y = act[j].second;
			bool xin = s[i][x];
			bool yin = s[i][y];
			if( xin &&  yin) s[i].reset();
			if( xin && !yin) s[i].set(y) ;
			if(!xin &&  yin) s[i].set(x) ;
		}
	}
	int ans = 0;
	for(int i = 1; i <= n; ++ i){
		for(int j = i + 1; j <= n; ++ j){
			if(s[i].any() && s[j].any()){
				std::bitset<MAXN> now = s[i] & s[j];
				ans += now.none();
			}
		}
	}
	printf("%d\n", ans);
	return 0;
}