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 N=410,M=1e5+10;
bitset<N> S[N];//S[i]表示让i活下来,哪些必死 
int n,m;
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		if (S[x][x]||S[y][y]||S[x][y]||S[y][x]||(S[x]&S[y]).any()) S[x][x]=S[y][y]=1;
		S[x]=(S[y]|=S[x]);
		S[x][y]=S[y][x]=1;
	}
	int cnt=0;
	for (int i=1;i<=n;i++)
	for (int j=1;j<i;j++)
	if (!S[i][i]&&!S[j][j]&&!S[i][j]&&!S[j][i]&&(S[i]&S[j]).none()) cnt++;
	printf("%d\n",cnt);
	return 0;
}