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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(ll i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;

using namespace std;
#define N 410
#define M 100010
ll n,m,x[M],y[M];
bool t[N][N],ok[N];
int main(){
    cin>>n>>m;
    for(ll i=m-1;~i;i--){
	cin>>x[i]>>y[i];
	x[i]--,y[i]--;
    }
    lol(i,n){
	lol(j,n)t[i][j]=0;
	t[i][i]=1;
	ok[i]=1;
	lol(j,m){
	    bool sx=t[i][x[j]],sy=t[i][y[j]];
	    if(sx==1&&sy==1)ok[i]=0;
	    if(sx==0&&sy==0)continue;
	    if(sx)t[i][y[j]]=1;
	    if(sy)t[i][x[j]]=1;
	}
    }
    ll ans=0;
    lol(i,n)lol(j,i){
	if(ok[i]==0||ok[j]==0)continue;
	bool cad=1;
	lol(k,n)if(t[i][k]&t[j][k])cad=0;
	ans+=cad;
    }
    cout<<ans<<endl;
    return 0;
}
