CodeForces

You are given three integers a, b and x. Your task is to construct a binary string s of length n=a+b such that there are exactly a zeroes, exactly b ones and exactly x indices i (where 1≤i<n) such that si≠si+1

. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices i

such that 1≤i<n and si≠si+1 (i=1,2,3,4). For the string "111001" there are two such indices i (i=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 a

, b and x (1≤a,b≤100,1≤x<a+b)

.

Output

Print only one string s

, where s

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<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b,x,n,i,j;
    char s[1010];
    cin>>a>>b>>x;
    n=a+b;
    if(a>b)
    {
        s[1]='0';
        a--;
        for(i=2;i<=x;i++)
        {
            if(s[i-1]=='0')
            {
                s[i]='1';
                b--;
            }
            else
            {
                 s[i]='0';
                 a--;
            }
        }
        if(s[x]=='0')
        {
            for(i=x+1;;i++)
            {
                if(a>0)
                {
                     s[i]='0';
                     a--;
                }
                else
                {
                    s[i]='1';
                    b--;
                }
                if(a==0&&b==0)
                    break;
            }
        }
        else
        {
            for(i=x+1;;i++)
            {
                if(b>0)
                {
                     s[i]='1';
                     b--;
                }
                else
                {
                    s[i]='0';
                    a--;
                }
                if(a==0&&b==0)
                    break;
            }
        }
    }