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 maxN=410;
const int maxM=101000;

int n,m;
int P1[maxM],P2[maxM],Mark[maxN];
bitset<maxN> B[maxN];

int main(){
    scanf("%d%d",&n,&m);for (int i=1;i<=m;i++) scanf("%d%d",&P1[i],&P2[i]);
    for (int i=1;i<=n;i++){
	B[i][i]=1;
	for (int j=m;j>=1;j--)
	    if (B[i][P1[j]]&&B[i][P2[j]]){
		Mark[i]=1;break;
	    }
	    else if (B[i][P1[j]]||B[i][P2[j]]) B[i][P1[j]]=B[i][P2[j]]=1;
    }
    int Ans=0;
    for (int i=1;i<=n;i++) for (int j=i+1;j<=n;j++) if (!Mark[i]&&!Mark[j]&&!(B[i]&B[j]).any()) ++Ans;
    printf("%d\n",Ans);return 0;
}
