Owl Geeks

 

The owls have the following equation:

Y = a × x2 + b × x

With a, b, and N given, they decide to put into a set the integer values of Y that are less than or equal to N and that are outputted from the equation from any positive integer x.

With that set of numbers, they come up with the problem of finding the winning digit among them.

The winning digit is a digit from 0 to 9 that will get the maximum number of points. How are points for a digit calculated you may ask? Well, be a bit more patient, I’m going to tell you now.

For each number in the set, if the digit was the most repeated digit or tied with other digits as the most repeated digit in the ith number of set S, then it would get one point from that ith number.

Can you tell the owls what the winning digit is?

Input

The first line of input is T – the number of test cases.

The first line of each test case is a, b, and N (1 ≤ a, b, N ≤ 105).

Output

For each test case, print on a line the winning digit with the maximum number of points. If there is a tie, print the minimum digit among them. If the set is empty, print  - 1.

Example

input

Copy

2
1 2 50
20 3 10

output

Copy

3
-1

题意:输入a,b,n,n是a*x*x+b*x的最大范围，例如第一组，n可以为3,8,15,24,35,48，将3,8,1,5,2,4,3,5,4,8存起来（要是332存3），输出最小而且个数最多的数

 

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int c[100],d[100];
int main()
{
    int T,a,b,n,r;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&a,&b,&n);
        r=0;
        memset(c,0,sizeof(c));
        memset(d,0,sizeof(d));
        for(int i=1;a*i*i+b*i<=n;i++)
               r=i;
        if(r==0)
        {
            puts("-1");
            continue;
        }
        for(int i=1;i<=r;i++)
        {
            int s=a*i*i+b*i,v=0;
            memset(c,0,sizeof(c));
            while(s)
            {
                c[s%10]++;
                v=max(v,c[s%10]);
                s/=10;
            }
            for(int j=0;j<=9;j++)
                if(c[j]==v)
                    d[j]++;
        }
        int u=0;
        for(int i=0;i<=9;i++)
            if(d[i]>d[u])
                u=i;
        printf("%d\n",u);
    }
    return 0;
}