CodeForces - 1003-B-Binary String Constructing (规律+模拟)

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 exactly xx 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 and si≠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

题解：

根据0和1谁的数量大，确定从谁开始，然后n为奇数和偶数再分别判断即可

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>

using namespace std;

int main()
{
    int a,b,x,s;
    int m,n;
    cin>>a>>b>>x;
    s=a+b;
    if(a>b)
	{
	m=0;
	n=1;
    }
    else
	{
	 m=1;
	 n=0;
	 swap(a,b);
	 }
    for(int i=0;i<x/2;i++) {cout<<m<<n;a--,b--;}
    if(x%2==0){
        while(b--) cout<<n;
        while(a--) cout<<m;
    }
    else{
        while(a--) cout<<m;
        while(b--) cout<<n;
    }
    return 0;
}