Magic Bracelet
 Time Limit: 2000MS Memory Limit: 131072K Total Submissions: 4990 Accepted: 1610

Description

Ginny’s birthday is coming soon. Harry Potter is preparing a birthday present for his new girlfriend. The present is a magic bracelet which consists of n magic beads. The are m kinds of different magic beads. Each kind of beads has its unique characteristic. Stringing many beads together a beautiful circular magic bracelet will be made. As Harry Potter’s friend Hermione has pointed out, beads of certain pairs of kinds will interact with each other and explode, Harry Potter must be very careful to make sure that beads of these pairs are not stringed next to each other.

There infinite beads of each kind. How many different bracelets can Harry make if repetitions produced by rotation around the center of the bracelet are neglected? Find the answer taken modulo 9973.

Input

The first line of the input contains the number of test cases.

Each test cases starts with a line containing three integers n (1 ≤ n ≤ 109gcd(n, 9973) = 1), m (1 ≤ m ≤ 10), k (1 ≤ k ≤ m(m − 1) ⁄ 2). The next k lines each contain two integers a and b (1 ≤ab ≤ m), indicating beads of kind a cannot be stringed to beads of kind b.

Output

Output the answer of each test case on a separate line.

Sample Input

4
3 2 0
3 2 1
1 2
3 2 2
1 1
1 2
3 2 3
1 1
1 2
2 2

Sample Output

4
2
1
0

Source

/*
poj 2888 Magic Bracelet(Polya+矩阵快速幂)

hhh-2016-04-19 22:31:42
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
//typedef long long ll;
using namespace std;
const int  maxn = ;
struct Matrix
{
int ma[][];
};

int n,m;
int isprime[maxn];
int prime[maxn];
int pnum;
void get_prime()
{
pnum = ;
memset(isprime, -, sizeof(isprime));
for(int i = ; i <= maxn-; i ++)
if(isprime[i])
{
prime[pnum ++] = i;
for(int j = i * i; j <= maxn-; j += i)
isprime[j] = ;
}
}

Matrix mult(Matrix ta,Matrix tb, int mod)
{
Matrix tc;
memset(tc.ma,,sizeof(tc.ma));
for(int i = ; i < m; i ++)
for(int k = ; k < m; k ++)
if(ta.ma[i][k])
{
for(int j = ; j < m; j ++)
if(tb.ma[k][j])
tc.ma[i][j] = (tc.ma[i][j] + ta.ma[i][k] * tb.ma[k][j]) % mod;
}
return tc;
}

Matrix Mat_pow(Matrix ta,int n,int mod)
{
Matrix t;
memset(t.ma,,sizeof(t.ma));
for(int i = ; i < m; i++)
t.ma[i][i] = ;
while(n)
{
if(n & ) t = mult(t,ta,mod);
ta = mult(ta,ta,mod);
n >>= ;
}
return t;
}

int pow_mod(int a,int n,int mod)
{
int ret = ;
a %= mod;
while(n)
{
if(n & ) ret = ret*a,ret%=mod;
a = a*a;
a%=mod;
n >>= ;
}
return ret;
}
int mod;
int euler(int cur )
{
int ans, x;
ans = x = cur;
for(int i = ; i < pnum && prime[i] * prime[i] <= cur; i++)
if(x % prime[i] == )
{
ans = ans / prime[i] * (prime[i] - );
while(x % prime[i] == )
x /= prime[i];
}
if(x > )
ans = ans / x * (x - );
return ans%mod;
}

Matrix mat;
int cal(int len)
{
int ret = ;
Matrix t = Mat_pow(mat,len,mod);
for(int i = ; i < m; i++)
ret = ret + t.ma[i][i];
ret %= mod;

return ret;
}

int Polya(int n)
{
int ans = ;
for(int i= ; i*i <= n; i++)
{
if(n % i == )
{
if(i*i == n)
{
ans = ans+cal(i)*euler(i);
ans%=mod;
}
else
{
ans = (ans+cal(i)*euler(n/i)+cal(n/i)*euler(i));
ans%=mod;
}
}
}
ans = ans*pow_mod(n,mod-,mod);
return ans % mod;
}

int main()
{
int T;
get_prime();
mod = ;
scanf("%d",&T);
while(T--)
{
int k;
scanf("%d%d%d",&n,&m,&k);
int u,v;
for(int i = ; i < m; i++)
for(int j = ; j < m; j++)
mat.ma[i][j] = ;
for(int i = ; i <= k; i++)
{
scanf("%d%d",&u,&v);
mat.ma[u-][v-] = mat.ma[v-][u-] = ;
}
printf("%d\n",Polya(n));
}
return ;
}

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