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<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#define N 410
#define M 100010
using namespace std;
struct bb{int x, y;}b[M];
int n, m, ans;
bitset<N>set[N];
int main(){
	scanf("%d%d", &n, &m);
	for(int i=1; i<=m; i++)scanf("%d%d", &b[i].x, &b[i].y);
	for(int i=1; i<=n; i++){
		set[i].reset(); set[i][i]=1;
		for(int j=m; j; j--){
			if(set[i][b[j].x]&&set[i][b[j].y]){set[i].reset(); break;}
			if(!set[i][b[j].x]&&!set[i][b[j].y])continue;
			set[i][b[j].x]=set[i][b[j].y]=1;
		}
	}
	ans=0;
	for(int i=1; i<=n; i++)if(set[i].count())
		for(int j=i+1; j<=n; j++)if(set[j].count())
			if(!(set[i]&set[j]).count())ans++;
	printf("%d", ans);
	return 0;
}