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 sz(c)      int(c.size())
#define rep(i,a,b) for (int i=a; i<(b); ++i)
#define per(i,a,b) for (int i=(b)-1; i>=(a); --i)
using namespace std;
using ll = long long;

int const MAXN=440;
int const MAXM=110000;
int N,M;
int x[MAXM],y[MAXM];
bool S[MAXN][MAXN];
bool can[MAXN];

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	cout<<fixed<<setprecision(10);

	cin>>N>>M;
	rep(i,0,M) cin>>x[i]>>y[i];
	
	rep(i,1,N+1) {
		rep(j,1,N+1) S[i][j]=false;
		S[i][i]=true;
		can[i]=true;
	
		per(j,0,M) {
			if (S[i][x[j]] && S[i][y[j]]) {
				can[i]=false;
				break;
			} else if (S[i][x[j]]) {
				S[i][y[j]]=true;
			} else if (S[i][y[j]]) {
				S[i][x[j]]=true;
			}
		}
	}
	
	int res=0;
	rep(i,1,N+1) rep(j,i+1,N+1) if (can[i] && can[j]) {
		bool canBoth=true;
		rep(k,1,N+1) if (S[i][k] && S[j][k]) {
			canBoth=false;
			break;
		}
		if (canBoth) res++;		
	}
	
	cout<<res<<"\n";
}
