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 std::bitset;

const int M=1e5+10,N=405;

int n,m,x[M],y[M];
bitset<N>q[N],P;

void Init() {
	scanf("%d%d",&n,&m);
	for (int i=0;i<m;++i) scanf("%d%d",&x[i],&y[i]),x[i]--,y[i]--;
}

void Solve() {
	for (int i=0;i<n;++i) P[i]=1;
	for (int i=0;i<n;++i) {
		q[i][i]=1;
		for (int j=m-1;j>=0;--j) {
			if (q[i][x[j]] && q[i][y[j]]) { q[i]=P; break; }
			if (q[i][x[j]]) q[i][y[j]]=1;
			else if (q[i][y[j]]) q[i][x[j]]=1;
		}
	}
	int cnt=0;
	for (int i=0;i<n;++i)
		for (int j=i+1;j<n;++j) 
			if ((q[i]&q[j]).count()==0) 
				cnt++;
	printf("%d\n",cnt);
}

int main() {
	Init();
	Solve();
	return 0;
}