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>
using namespace std;
typedef pair<int,int>pii;
const int N=405,M=1e5+5;
int n,m,x[M],y[M],i,j,k,a[N],aa[N][N],ans;
bool b[N],bb[M];
vector<pii>e[N];
void dfs(int u){
	for(pii v:e[u])
		if(!a[v.first] && v.second<a[u])
			bb[a[v.first]=v.second]=1,dfs(v.first);
}
int main(){
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;++i)scanf("%d%d",x+i,y+i);
	for(i=m;i;--i)e[x[i]].push_back(pii(y[i],i)),e[y[i]].push_back(pii(x[i],i));
	for(i=1;i<=n;++i){
		memset(bb+1,0,m);memset(a+1,0,n<<2);a[i]=M;
		dfs(i);
		for(j=1;j<=m;++j)if(!bb[j] && a[x[j]] && a[y[j]] && (a[x[j]]>j || a[y[j]]>j))break;
		if(j<=m)b[i]=1;
		memcpy(aa[i]+1,a+1,n<<2);
	}
	for(i=1;i<n;++i)if(!b[i])for(j=i+1;j<=n;++j)if(!b[j]){
		for(k=1;k<=n;++k)if(aa[i][k] && aa[j][k] && aa[i][k]!=aa[j][k])break;
		ans+=k>n;
	}
	printf("%d\n",ans);
	return 0;
}