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>
const int N=1e5+7;
int n,m,e[N][2],ans=0;
bool S[407][407];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;++i)scanf("%d%d",e[i],e[i]+1);
	for(int i=1;i<=n;++i){
		S[i][0]=S[i][i]=1;
		for(int j=m;j;--j){
			int x=e[j][0],y=e[j][1];
			if(S[i][x]){
				if(S[i][y]){
					S[i][0]=0;
					break;
				}
				S[i][y]=1;
			}else if(S[i][y])S[i][x]=1;
		}
	}
	for(int i=1;i<=n;++i)if(S[i][0]){
		for(int j=1;j<i;++j)if(S[j][0]){
			++ans;
			for(int k=1;k<=n;++k)if(S[i][k]&S[j][k]){
				--ans;
				break;
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}