Binary String Constructing

                                                                          time limit per test 1 second

                                                                   memory limit per test 256 megabytes

                                                                               input standard input

                                                                             output standard output

You are given three integers aa, bb and xx. Your task is to construct a binary string ss of 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

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.

题解

题目大致意思是：a个0，b个1，组成一个长为a+b的字符串s，要求有x个位置使得si≠si+1

因为题目不要求格式，只要符合题意就可以。所以可以先放多的，也就是a大先放0，b大先放1，交替放置。需要注意的是，当x为1时要一次性把剩下的0或1全部放置，具体放0还是1，可以设一个标记mark

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

using namespace std;
const int maxn = 200+ 10;

int main()
{
	int a, b, x;
	scanf("%d%d%d",&a,&b,&x);
	int mark;
	if(a>=b) mark = 0;
	else  mark = 1;
	while(x)
	{
		if(x==1) 
		{
			if(mark) 
			{
				for(int i=0;i<b;i++) printf("1");
				for(int i=0;i<a;i++) printf("0");
			}
			
			else 
			{
				for(int i=0;i<a;i++) printf("0");
				for(int i=0;i<b;i++) printf("1");
			}
		}
		else
		{
			printf("%d",mark);
			if(mark) 
			{
				b--;
				mark = 0;
			}
			else 
			{
				a--;
				mark = 1;
			}
		}
		x--;
	}
	
	
	printf("\n");
	return 0;
 }