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 il inline
#define vd void
typedef long long ll;
il ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
int a[100010],b[100010];
std::bitset<410>bit[410];
bool del[410];
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),m=gi();
	for(int i=1;i<=m;++i)a[i]=gi(),b[i]=gi();
	for(int i=1;i<=n;++i){
		bit[i][i]=1;
		for(int j=m;j;--j){
			bool x=bit[i][a[j]],y=bit[i][b[j]];
			if(x&&y){del[i]=1;break;}
			if(x)bit[i][b[j]]=1;
			if(y)bit[i][a[j]]=1;
		}
	}
	int ans=0;
	for(int i=1;i<=n;++i){
		if(del[i])continue;
		for(int j=i+1;j<=n;++j){
			if(del[j])continue;
			if(!(bit[i]&bit[j]).any())++ans;
		}
	}
	printf("%d\n",ans);
	return 0;
}
