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>
typedef unsigned long long ull;
const int MaxN=405,MaxM=100005;
int N,M,x[MaxM],y[MaxM];
ull g[MaxN][MaxN/64+1],del[MaxN/64+1],v[MaxN][MaxN/64+1];
int main(){
	scanf("%d%d",&N,&M);
	for(int i=0;i<M;i++)scanf("%d%d",x+i,y+i),--x[i],--y[i];
	for(int i=0;i<N;i++)g[i][i>>6]|=1ull<<(i&63);
	for(int i=M;i--;){
		ull*gx=g[x[i]],*gy=g[y[i]];
		for(int j=0;j<<6<N;j++){
			del[j]|=gx[j]&gy[j];
			gx[j]=gy[j]=gx[j]|gy[j];
		}
	}
	int cnt=0,ans=0;
	for(int i=0;i<N;i++)if(!(del[i>>6]>>(i&63)&1)){
		for(int j=0;j<N;j++)if(g[j][i>>6]>>(i&63)&1)v[cnt][j>>6]|=1ull<<(j&63);
		for(int j=0;j<cnt;j++){
			bool fl=1;
			for(int k=0;k<<6<N&&fl;k++)if(v[j][k]&v[cnt][k])fl=0;
			if(fl)ans++;
		}
		cnt++;
	}
	printf("%d\n",ans);
}