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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>

using namespace std;
typedef long long LL;
const int maxn = 405;
const int maxm = 100005;
const int inf = 1<<29;

int x[maxm],y[maxm],n,m,ans;
int bac[maxn][maxn],ok[maxn];
int main()
{
	#ifdef Amberframe
		freopen("agc016e.in","r",stdin);
		freopen("agc016e.out","w",stdout);
	#endif
	scanf("%d %d",&n,&m);
	for (int i=1;i<=m;i++)
		scanf("%d %d",&x[i],&y[i]);
	for (int i=1;i<=n;i++) {
		bac[i][i]=1;ok[i]=true;
		for (int j=m;j>=1;j--)
		if (bac[i][x[j]]&&bac[i][y[j]]) {ok[i]=false;break;}
		else if (bac[i][x[j]]) bac[i][y[j]]=1;
		else if (bac[i][y[j]]) bac[i][x[j]]=1;
	}
	
	for (int i=1;i<=n;i++)
	for (int j=i+1;j<=n;j++)
	if (ok[i]&&ok[j]) {
		int cnt=0;
		for (int k=1;k<=n;k++)
			cnt+=bac[i][k]&&bac[j][k];
		ans+=(cnt==0);
	}
	printf("%d",ans);
	return 0;
}