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 re register
#define inc(i,j,k) for(re int i=j;i<=k;i++)
#define dec(i,j,k) for(re int i=j;i>=k;i--) 
using namespace std;
const int N=405;
const int M=1e5+5;
inline int read(){
	int x=0;
	char ch=getchar();
	while(ch<'0'||ch>'9') ch=getchar();
	while('0'<=ch&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x;
}
int n,m,a[M],b[M],ans;
bool dp[N][N],die[N];
int main(){
	n=read(),m=read();
	inc(i,1,m) a[i]=read(),b[i]=read();
	inc(i,1,n){
		dp[i][i]=1;
		dec(j,m,1){
			bool x=dp[i][a[j]],y=dp[i][b[j]];
			if(x&&y){die[i]=1;break;}
			if(x) dp[i][b[j]]=1;
			if(y) dp[i][a[j]]=1;
		}
	}
	inc(i,1,n){
		if(die[i]) continue;
		inc(j,i+1,n){
			if(die[j]) continue;
			ans++;
			inc(k,1,n){
				if(dp[i][k]&&dp[j][k]){
					ans--;
					break;
				}
			} 
		}
	}
	printf("%d\n",ans);
}