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 F first
#define S second
#define X real()
#define Y imag()
using namespace std;
typedef long long ll;
typedef long double ld;

int n,m;

vector<pair<int, int> > g[444];

int kt[444];

int fail=0;

void killl(int x, int t){
	if (fail) return;
	if (kt[x]){
		fail=1;
		return;
	}
	kt[x]=t;
	for (auto nx:g[x]){
		if (nx.S>=t) break;
		killl(nx.F, nx.S);
	}
}

int test(int a, int b){
	for (int i=1;i<=n;i++){
		kt[i]=0;
	}
	fail=0;
	killl(a, m+1);
	killl(b, m+1);
	return fail==0;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin>>n>>m;
	for (int i=1;i<=m;i++){
		int x,y;
		cin>>x>>y;
		g[x].push_back({y, i});
		g[y].push_back({x, i});
	}
	int v=0;
	for (int i=1;i<=n;i++){
		for (int ii=i+1;ii<=n;ii++){
			v+=test(i, ii);
		}
	}
	cout<<v<<endl;
}