#include <bits/stdc++.h>
#define Mid ((l + r) >> 1)
#define lson (rt << 1)
#define rson (rt << 1 | 1)
using namespace std;
int read(){
	char c; int num, f = 1;
	while(c = getchar(),!isdigit(c)) if(c == '-') f = -1; num = c - '0';
	while(c = getchar(), isdigit(c)) num = num * 10 + c - '0';
	return f * num;
}
const int N = 2e5 + 1009;
int a, b, k;
int x[N], y[N];
signed main()
{
	a = read(); b = read(); k = read();
	if(k == 0) {
		printf("Yes\n");
		for(int i = 1; i <= b; i++) printf("1");
		for(int i = 1; i <= a; i++) printf("0");printf("\n");
		for(int i = 1; i <= b; i++) printf("1");
		for(int i = 1; i <= a; i++) printf("0");printf("\n");
		return 0;
	}
	if(a + b < k + 2 || (a > 0 && b == 1) || a == 0) {
		printf("No\n");
		return 0;
	}
	if(a >= k) {
		y[1] = 1;
		x[k + 1] = 1;
		for(int i = a + 2; i <= a + b; i++)
			x[i] = y[i] = 1;
	} else {
		y[1] = 1;
		x[k + 1] = 1;
		for(int i = a + 1; i <= k; i++) {
			x[i] = y[i] = 1;
		}
		for(int i = k + 2; i <= a + b; i++)
			x[i] = y[i] = 1;
	}
	reverse(x + 1, x + 1 + a + b);
	reverse(y + 1, y + 1 + a + b);
	printf("Yes\n");
	for(int i = 1; i <= a + b; i++) printf("%d", x[i]); printf("\n");
	for(int i = 1; i <= a + b; i++) printf("%d", y[i]); printf("\n");
	return 0;
}
/*
两个位置都为1
表示这两个位置之间的数都变成1
也就是说只用两个1就可以构造出来了

可以把中间一对0换成1
可以减少0的数量
至少需要2+k
*/