#include <bits/stdc++.h> //常规版
using namespace std;
const int N = 1010;
char s[N];
int a[N], b[N], cti[150], itc[150];
int main()
{
    for (char i = '0'; i <= '9'; i++) cti[i] = i - '0', itc[i - '0'] = i;
    for (char i = 'A'; i <= 'Z'; i++) cti[i] = i - 'A' + 10, itc[i - 'A' + 10] = i;
    for (char i = 'a'; i <= 'z'; i++) cti[i] = i - 'a' + 36, itc[i - 'a' + 36] = i;
    int n, m;
    while (scanf("%d%s%d", &n, s + 1, &m) != EOF)
    {

        int len = strlen(s + 1);
        for (int i = 1; i <= len; i++) a[len - i + 1] = cti[s[i]];
        int slen = 0;
        while (len)
        {
            int x = 0;
            for (int i = len; i >= 1; i--)
            {
                int t = x * n + a[i];
                x = t % m;
                a[i] = t / m;
            }
            while (!a[len] && len >= 1) len--;
            s[++slen] = itc[x];
        }
        for (int i = slen; i >= 1; i--) printf("%c", s[i]);
        puts("");
    }
    return 0;
}

#include <bits/stdc++.h> //高精度版
using namespace std;
typedef long long LL;
const int N = 1010;
char s1[N], s2[N];
int cti[150], itc[150];
struct node
{
    int len, a[N];
    node()
    {
        len = 1;
        memset(a, 0, sizeof(a));
    }
};
node operator+(node n1, int x)
{
    node no; no = n1;
    no.a[1] += x;
    for (int i = 1; i <= no.len; i++)
    {
        no.a[i + 1] += no.a[i] / 10;
        no.a[i] %= 10;
    }
    int i = no.len;
    while (no.a[i + 1] > 0)
    {
        i++;
        no.a[i + 1] += no.a[i] / 10;
        no.a[i] %= 10;
    }
    while (i > 1 && no.a[i] == 0)
        i--;
    no.len = i;
    return no;
}
node operator*(node n1, int x)
{
    node no; no.len = n1.len;
    for (int i = 1; i <= no.len; i++) no.a[i] = n1.a[i] * x;
    for (int i = 1; i <= no.len; i++)
    {
        no.a[i + 1] += no.a[i] / 10;
        no.a[i] %= 10;
    }
    int i = no.len;
    while (no.a[i + 1] > 0)
    {
        i++;
        no.a[i + 1] += no.a[i] / 10;
        no.a[i] %= 10;
    }
    while (i > 1 && no.a[i] == 0)
        i--;
    no.len = i;
    return no;
}
int tx;
node operator/(node n1, int x)
{
    node no;
    int t = 0;
    no.len = n1.len;
    for (int i = n1.len; i >= 1; i--)
    {
        t = t * 10 + n1.a[i];
        no.a[i] = t / x;
        t = t % x;
    }
    int i = no.len;
    while (i > 1 && no.a[i] == 0) i--;
    no.len = i;
    tx = t;
    return no;
}
int main()
{
    for (char c = 'A'; c <= 'Z'; c++) cti[c] = c - 'A' + 10, itc[c - 'A' + 10] = c;
    for (char c = 'a'; c <= 'z'; c++) cti[c] = c - 'a' + 36, itc[c - 'a' + 36] = c;
    for (char c = '0'; c <= '9'; c++) cti[c] = c - '0', itc[c - '0'] = c;
    int n, m;
    while (scanf("%d%s%d", &n, s1 + 1, &m) != EOF)
    {
        node no;
        int len1 = strlen(s1 + 1);
        for (int i = 1; i <= len1; i++)
        {
            no = no * n;
            no = no + cti[s1[i]];
        }
        int len2 = 0;
        while (!(no.len == 1 && no.a[1] == 0))
        {
            no = no / m;
            s2[++len2] = tx;
        }
        for (int i = len2; i >= 1; i--) printf("%c", itc[s2[i]]);
        printf("\n");
    }
    return 0;
}