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;
#define Set(a,b) memset(a,b,sizeof(a))
template<class T>inline void init(T&x){
	x=0;char ch=getchar();bool t=0;
	for(;ch>'9'||ch<'0';ch=getchar()) if(ch=='-') t=1;
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+(ch-48);
	if(t) x=-x;return;
}typedef long long ll;
const int N=401;
const int M=1e5+10;
int x[M],y[M];
int n,m;
bitset<N>del[N];
bool cant[N];
int main()
{
	init(n),init(m);int u,v;
	for(int i=1;i<=m;++i) {
		init(u),init(v);
		if(u>v) swap(u,v);
		x[i]=u,y[i]=v;
	}
	for(int u=1;u<=n;++u) {
		del[u].set(u);
		for(int i=m;i;--i) {
			int a=x[i],b=y[i];
			if(del[u][a]&&del[u][b]) {cant[u]=1;break;}
			if(del[u][a]) del[u].set(b);
			if(del[u][b]) del[u].set(a);
		}
	}int ans=0;
	for(int i=1;i<=n;++i) {
		if(cant[i]) continue;
		for(int j=i+1;j<=n;++j){
			if(cant[j]) continue;
			if(!(del[i]&del[j]).any()) ++ans;
		}
	}cout<<ans<<endl;
	return 0;
}
