HDU ACM 3443

#include<iostream>
#include<cmath>
using namespace std;

int main()
{
        long long n, m, sum;
        while(cin >> n, n){
                int count = 0;
                m = n;
                while(m)
                {
                        count++;
                        m /= 10;
                }
                sum = (long long)(pow(10.0, 1.0 * count) - 1) / 9;//1 + 10 + 100 + 1000 + 10000……的公式
                while(n % sum != 0){
                        sum /= 10;
                }
                cout << n / sum << endl;
        }
        return 0;
}