//给一个无向图 , 每条边能够是online边也能够是offline边,问
//有多少种方法使得每一个节点的online边和offline边一样多
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 110 ;
int on[maxn] , off[maxn];
int x[maxn] , y[maxn] ;
int degree[maxn] ;
int ans ;
int n , m ;
void dfs(int num)
{
if(num == m+1)
{
for(int i = 1;i <= n;i++)
if(on[i] != off[i])
return ;
ans++;return ;
}
int u = x[num] , v = y[num] ;
if((on[u] < degree[u]/2) && (on[v] < degree[v]/2))
{
on[u]++;on[v]++;
dfs(num+1);
on[u]--;on[v]--;
}
if(off[v] < degree[v]/2 && off[u] < degree[u]/2)
{
off[u]++;off[v]++ ;
dfs(num+1) ;
off[u]--;off[v]-- ;
}
}
int main()
{
int T ;
scanf("%d" ,&T) ;
while(T--)
{
scanf("%d%d" ,&n , &m) ;
memset(degree , 0 , sizeof(degree)) ;
memset(on , 0 , sizeof(on)) ;
memset(off, 0 , sizeof(off)) ;
for(int i = 1;i <= m;i++)
{
scanf("%d%d" ,&x[i] ,&y[i]) ;
degree[x[i]]++;
degree[y[i]]++;
}
ans = 0 ;
dfs(1);
printf("%d " , ans) ;
}
return 0 ;
}
//有多少种方法使得每一个节点的online边和offline边一样多
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 110 ;
int on[maxn] , off[maxn];
int x[maxn] , y[maxn] ;
int degree[maxn] ;
int ans ;
int n , m ;
void dfs(int num)
{
if(num == m+1)
{
for(int i = 1;i <= n;i++)
if(on[i] != off[i])
return ;
ans++;return ;
}
int u = x[num] , v = y[num] ;
if((on[u] < degree[u]/2) && (on[v] < degree[v]/2))
{
on[u]++;on[v]++;
dfs(num+1);
on[u]--;on[v]--;
}
if(off[v] < degree[v]/2 && off[u] < degree[u]/2)
{
off[u]++;off[v]++ ;
dfs(num+1) ;
off[u]--;off[v]-- ;
}
}
int main()
{
int T ;
scanf("%d" ,&T) ;
while(T--)
{
scanf("%d%d" ,&n , &m) ;
memset(degree , 0 , sizeof(degree)) ;
memset(on , 0 , sizeof(on)) ;
memset(off, 0 , sizeof(off)) ;
for(int i = 1;i <= m;i++)
{
scanf("%d%d" ,&x[i] ,&y[i]) ;
degree[x[i]]++;
degree[y[i]]++;
}
ans = 0 ;
dfs(1);
printf("%d " , ans) ;
}
return 0 ;
}