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;
#define R register

inline void read(int &x){
	static int ch; static bool flag;
	for(flag=0,ch=getchar(); ch>'9'||ch<'0'; ch=getchar()) flag |= ch=='-';
	for(x=0; ch>='0'&&ch<='9'; ch=getchar()) x = (x<<1) + (x<<3) + ch-48;
	x = flag ? -x : x;
}

bool die[402];
bitset<402>s[402];
int N,M,Ans,x[100002],y[100002];

int main()
{
	read(N),read(M);
	for(R int i=1;i<=M;++i) read(x[i]),read(y[i]);
	for(R int i=1;i<=N;++i){
		s[i][i]=1;
		for(R int j=M;j;--j)
			if(s[i][x[j]] && s[i][y[j]]){die[i]=1;break;}
		else if(s[i][x[j]]) s[i][y[j]]=1;
		else if(s[i][y[j]]) s[i][x[j]]=1;
	}
	for(R int i=1;i<N;++i)
		if(!die[i])
			for(R int j=i+1;j<=N;++j)
				if(!die[j])
					if((s[i]&s[j]).none()) Ans++;
	printf("%d\n",Ans);
	return 0;
}
