tjut 3609

#include <iostream>  
using namespace std;  
#define LL unsigned __int64  
#define M 205  
  
int phi[M];  
  
int Euler (int n)    //求n的歐拉值  
{  
    int i, res = n;  
    for (i = 2; i * i <= n; i++)  
    {  
        if (n % i == 0)  
        {  
            res = res - res/i;  
            while (n % i == 0)  
                n /= i;  
        }  
    }  
    if (n > 1)  
        res = res - res/n;  
    return res;  
}  
  
LL qmod (LL a, LL b, int c)    //快速冪取模  
{  
    LL res = 1;  
    for ( ; b; b >>= 1)  
    {  
        if (b & 1)  
            res = res * a % c;  
        a = a * a % c;  
    }  
    return res;  
}  
  
LL isok (LL a, LL b, int c)    //目的是判斷a^b是否>=c  
{  
    LL res = 1, i;  
    for (i = 0; i < b; i++)  
    {  
        res *= a;  
        if (res >= c)  
            return res;  
    }  
    return res;  
}  
  
LL upup (LL a, int k, int num)    //按照題意遞歸地使用公式  
{  
    if (phi[num] == 1) return 1;    //顯然符合公式條件，很容易套公式知道是1，這實際上是剪枝  
    if (k == 1) return a % phi[num];//返回上一層的a的冪  
    LL b = upup (a, k-1, num+1);    //得到a的冪b  
    LL x = isok (a, b, phi[num]);  
    if (x >= phi[num])    //使用公式的條件，滿足條件可以進行降冪，返回的是上層a的冪  
        return qmod (a % phi[num], b, phi[num]) + phi[num];  
    else return x;    //不使用公式，直接返回a^b，也屬於上層a的冪  
}  
  
int main()  
{  
    LL a;  
    int k;  
    phi[0] = 100000000;  
    for (k = 1; k < M; k++)  
        phi[k] = Euler (phi[k-1]);    //預處理所需歐拉值  
    while (~scanf ("%I64u%d", &a, &k))  
    {  
        printf ("%I64u\n", upup (a, k, 0) % phi[0]);  
    }  
    return 0;  
}