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 FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;

const int MAX_N=400,MAX_M=1e5;

int N,M,x[MAX_M],y[MAX_M];

bool flg[MAX_N];
bool flg2[MAX_N][MAX_N];

int main(){
	cin>>N>>M;
	REP(i,M){
		cin>>x[i]>>y[i];
		x[i]--;
		y[i]--;
	}
	REP(i,N){
		flg2[i][i]=true;
		REP(j,M){
			bool &f1=flg2[i][x[M-1-j]],&f2=flg2[i][y[M-1-j]];
			if (f1){
				if (f2){
					flg[i]=true;
					break;
				}else{
					f2=true;
				}
			}else{
				if (f2){
					f1=true;
				}
			}
		}
	}
	int ans=0;
	REP(i,N){
		if (flg[i]){
			continue;
		}
		FOR(j,i+1,N){
			if (flg[j]){
				continue;
			}
			bool f=false;
			REP(k,N){
				if (flg2[i][k] && flg2[j][k]){
					f=true;
					break;
				}
			}
			if (!f){
				ans++;
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}