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<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;

const int maxn = 510;
const int maxm = 210000;

int n,m;
struct edge{int x,y;}a[maxm];
bitset<maxn>S[maxn];

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++) scanf("%d%d",&a[i].x,&a[i].y);
	
	for(int i=1;i<=n;i++)
	{
		S[i][i]=1;
		for(int j=m;j>=1;j--)
		{
			int x=a[j].x,y=a[j].y;
			if(S[i][x]&&S[i][y]) { S[i].set(); break; }
			if(!S[i][x]&&!S[i][y]) continue;
			if(S[i][x]) S[i][y]=1;
			else S[i][x]=1;
		}
	}
	
	int ans=0;
	for(int i=1;i<n;i++) for(int j=i+1;j<=n;j++) if((S[i]&S[j]).count()==0)
		ans++;
	printf("%d\n",ans);
	
	return 0;
}
