#include<cstring>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
string s;
int m, k; int i, j; int a[205], aans[205], n[205], ans[205], last[205], now[205], t[205];
int single_j[12] = { 1,1,4,4,2,1,1,4,4,2 };//单循环结
void init() {
    memset(a, 0, sizeof a); memset(last, 0, sizeof last);
    memset(aans, 0, sizeof aans); memset(now, 0, sizeof now);
    memset(ans, 0, sizeof ans); memset(n, 0, sizeof n); memset(t, 0, sizeof t);
    //for (int i = 1;; i++) { if (m == 0) break; n[i] = m % 10; m /= 10; }//将m存入数组n，以便于高精度
}
void multiplyh(int x[], int y[], int z[]) {//高精度高乘
    int up = 0;
    for (int ii = 1; ii <= k; ii++) {
        for (j = 1; j <= k; j++)
        {
            z[ii + j - 1] += (x[j] * y[ii] + up) % 10;
            up = (x[j] * y[ii] + up) / 10;
        }
        up = 0;
    }
    for (int ii = 1; ii <= k; ii++) {//进位
        z[ii + 1] += z[ii] / 10;
        z[ii] %= 10;
    }
}
void multiplyl(int x[], int yy, int z[]) {//高精度低乘
    int up = 0;
    for (int ii = 1; ii <= k; ii++) {
        z[ii] = (x[ii] * yy + up) % 10;
        up = (x[ii] * yy + up) / 10;
    }
}
int main()
{
    //scanf("%d%d", &m, &k);
    init();
    cin >> s;
    cin >> k;
    int temp = 0, len = s.size();
    for (i = len - 1; i >= len - k; i--)
        n[++temp] = s[i] - '0';
    int tmp = 0;
    for (int i = 1; i <= k; i++) ans[i] = n[i];
    for (int i = 1; i < single_j[n[1]]; i++) {
        memset(aans, 0, sizeof aans);
        multiplyh(ans, n, aans);
        for (int j = 1; j <= k; j++) { ans[j] = aans[j]; }//更新为第一次出现末尾循环节的状态
    }
    t[1] = single_j[n[1]];//最低位的循环结
    for (int i = 1; i <= k; i++) now[i] = ans[i];
    int pos = 2;//当前倒数位数
    while (pos <= k) {
        for (int j = 1; j <= k; j++) {
            ans[j] = n[j];
            last[j] = now[j];
        }
        tmp = 0;
        while (tmp < 11) {
            tmp++;
            memset(aans, 0, sizeof aans);
            multiplyh(ans, now, aans);
            for (j = 1; j <= k; j++) {
                ans[j] = aans[j];
            }
            if (ans[pos] == n[pos]) break;//找到循环结
            memset(aans, 0, sizeof(aans));
            multiplyh(last, now, aans);//更新last
            for (j = 1; j <= k; j++) last[j] = aans[j];
        }
        if (tmp >= 11) { cout << -1; return 0; }
        for (int j = 1; j <= k; j++) now[j] = last[j];
        memset(aans, 0, sizeof aans);
        multiplyl(t, tmp, aans);//更新循环节数组
        for (int i = 1; i <= 100; i++)  t[i] = aans[i];
        pos++;
    }
    int flag = 0;//不输出前导0
    for (int i = 100; i >= 1; i--) {
        if (t[i]) flag = 1;
        if (flag) cout << t[i];
    }
    //cout << endl;
    return 0;
}