Binary String Constructing CodeForces - 1003B

You are given three integers aa, bb and xx. Your task is to construct a binary string ssof length n=a+bn=a+b such that there are exactly aa zeroes, exactly bb ones and exactlyxx indices ii (where 1≤i<n1≤i<n) such that si≠si+1si≠si+1. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices ii such that 1≤i<n1≤i<n andsi≠si+1si≠si+1 (i=1,2,3,4i=1,2,3,4). For the string "111001" there are two such indices ii (i=3,5i=3,5).

Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.

Input

The first line of the input contains three integers aa, bb and xx (1≤a,b≤100,1≤x<a+b)1≤a,b≤100,1≤x<a+b).

Output

Print only one string ss, where ss is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.

Examples

Input

2 2 1

Output

1100

Input

3 3 3

Output

101100

Input

5 3 6

Output

01010100

Note

All possible answers for the first example:

• 1100;
• 0011.

All possible answers for the second example:

• 110100;
• 101100;
• 110010;
• 100110;
• 011001;
• 001101;
• 010011;

• 001011.

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int a,b,n,x,i,j,minn;
    char c,d;
    minn = 10000;
    cin>>a>>b>>n;
    minn = min(a,b);
    if(a>b) c = '0',d = '1';
    else c = '1',d = '0';
    if(n%2==0)
    {
        for(i=0;i<n;i++)
        {
            if(i%2==0) printf("%c",c);
            else printf("%c",d);
        }
        //cout<<i<<minn<<minn-n/2<<endl;
        for(;i<n+minn-n/2;i++)
            printf("%c",d);
        for(;i<a+b;i++)
        {
            printf("%c",c);
        }
    }
    else
    {
        for(i=0;i<n;i++)
        {
            if(i%2==0) printf("%c",d);
            else printf("%c",c);
        }
        for(;i<n+minn-n/2-1;i++)
            printf("%c",d);
        for(;i<a+b;i++)
        {
            printf("%c",c);
        }
    }
    return 0;
}