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;

const int MaxM=1e5+5;

int N,M,Ans;
int X[MaxM],Y[MaxM];
int Lst[405],T[405][405];
bool Tag[405];

int main(){
	int i,j,k;
	bool f;
	scanf("%d%d",&N,&M);
	for(i=1;i<=M;i++){
		scanf("%d%d",&X[i],&Y[i]);
		Lst[X[i]]=Lst[Y[i]]=i;
	}
	for(i=1;i<=N;i++){
		Tag[i]=true,T[i][i]=Lst[i]+1;
		for(j=Lst[i];j;j--){
			if(T[i][X[j]]&&T[i][Y[j]]){
				Tag[i]=false;
				break;
			}
			if(T[i][X[j]])
				T[i][Y[j]]=j;
			else if(T[i][Y[j]])
				T[i][X[j]]=j;
		}
	}
	for(i=1;i<N;i++)
		for(j=i+1;j<=N;j++)
			if(Tag[i]&&Tag[j]){
				if(T[j][i]&&T[j][i]<=Lst[i])
					continue;
				if(T[i][j]&&T[i][j]<=Lst[j])
					continue;
				f=true;
				for(k=1;k<=N;k++)
					if(k^i&&k^j&&T[i][k]&&T[j][k]&&T[i][k]^T[j][k]){
						f=false;
						break;
					}
				Ans+=f;
			}
	printf("%d\n",Ans);
	return 0;
}