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;
int n;
bool g[444][444];
int main(){
	int m;
	scanf("%d%d",&n,&m);
	for(int i=1; i<=n; i++)
	for(int j=1; j<=n; j++)if(i!=j)
		g[i][j] = true;
	while(m--){
		int x,y;
		scanf("%d%d",&x,&y);
		if(g[x][y]){
			g[x][y] = g[y][x] = false;
			for(int i=1; i<=n; i++)if(x!=i)
				if(!g[x][i])
					g[i][y] = g[y][i] = false;
			for(int i=1; i<=n; i++)if(y!=i)
				if(!g[y][i])
					g[i][x] = g[x][i] = false;
		}else
		for(int i=1; i<=n; i++)
			g[x][i] = g[i][x] = g[y][i] = g[i][y] = false;
	}
	int res=0;
	for(int i=1; i<=n; i++)
		for(int j=i+1; j<=n; j++)
			if(g[i][j]){
				res++;
			}
	printf("%d\n",res);
	return 0;
}
