#include<bits/stdc++.h>
#define int long long
#define N 300000
using namespace std;
int a,b,k;
signed main(){
	cin>>a>>b>>k;b--;
	if(k==0){ 
		cout<<"Yes"<<endl;
		for(int i=1;i<=b+1;i++)printf("1");for(int i=1;i<=a;i++)printf("0");printf("\n");
		for(int i=1;i<=b+1;i++)printf("1");for(int i=1;i<=a;i++)printf("0");printf("\n");
		return 0;
	}
	if(k>a+b-1||!a||!b)cout<<"No"<<endl;
	else{
		cout<<"Yes"<<endl;
		if(k<=b){
			printf("1");
			for(int i=1;i<=b;i++)printf("1");for(int i=1;i<=a;i++)printf("0");printf("\n");
			printf("1");
			for(int i=1;i<=b-k;i++)printf("1");
			printf("0");
			for(int i=1;i<=k;i++)printf("1");
			for(int i=1;i<a;i++)printf("0");
		}else if(k<=a){
			printf("1");
			for(int i=1;i<=b;i++)printf("1");
			for(int i=1;i<=a;i++)printf("0");
			printf("\n");
			printf("1");
			for(int i=1;i<=k;i++)printf("0");
			for(int i=1;i<=b;i++)printf("1");
			for(int i=1;i<=a-k;i++)printf("0");
			printf("\n");
		}else{
			printf("1");
			for(int i=1;i<=b;i++)printf("1");
			for(int i=1;i<=a;i++)printf("0");
			printf("\n");
			printf("1");
			printf("0");
			for(int i=1;i<b;i++)printf("1");
			for(int i=1;i<a-(a+b-1-k);i++)printf("0");
			printf("1");
			for(int i=1;i<=(a+b-1-k);i++)printf("0");
			printf("\n");
		}
		printf("\n");
	}
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

You are given three integers a, b, k.

Find two binary integers x and y (x ≥ y) such that 

  1. both x and y consist of a zeroes and b ones; 
  2. x - y (also written in binary form) has exactly k ones. 

You are not allowed to use leading zeros for x and y. 

Input

The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.

Output

If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.

Otherwise print "No".

If there are multiple possible answers, print any of them.

Examples

Input


4 2 3


Output


Yes
101000
100001


Input


3 2 1


Output


Yes
10100
10010


Input


3 2 5


Output


No

Note

In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.

In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.

In the third example, one may show, that it's impossible to find an answer.

### Response

```cpp
#include<bits/stdc++.h>
#define int long long
#define N 300000
using namespace std;
int a,b,k;
signed main(){
	cin>>a>>b>>k;b--;
	if(k==0){ 
		cout<<"Yes"<<endl;
		for(int i=1;i<=b+1;i++)printf("1");for(int i=1;i<=a;i++)printf("0");printf("\n");
		for(int i=1;i<=b+1;i++)printf("1");for(int i=1;i<=a;i++)printf("0");printf("\n");
		return 0;
	}
	if(k>a+b-1||!a||!b)cout<<"No"<<endl;
	else{
		cout<<"Yes"<<endl;
		if(k<=b){
			printf("1");
			for(int i=1;i<=b;i++)printf("1");for(int i=1;i<=a;i++)printf("0");printf("\n");
			printf("1");
			for(int i=1;i<=b-k;i++)printf("1");
			printf("0");
			for(int i=1;i<=k;i++)printf("1");
			for(int i=1;i<a;i++)printf("0");
		}else if(k<=a){
			printf("1");
			for(int i=1;i<=b;i++)printf("1");
			for(int i=1;i<=a;i++)printf("0");
			printf("\n");
			printf("1");
			for(int i=1;i<=k;i++)printf("0");
			for(int i=1;i<=b;i++)printf("1");
			for(int i=1;i<=a-k;i++)printf("0");
			printf("\n");
		}else{
			printf("1");
			for(int i=1;i<=b;i++)printf("1");
			for(int i=1;i<=a;i++)printf("0");
			printf("\n");
			printf("1");
			printf("0");
			for(int i=1;i<b;i++)printf("1");
			for(int i=1;i<a-(a+b-1-k);i++)printf("0");
			printf("1");
			for(int i=1;i<=(a+b-1-k);i++)printf("0");
			printf("\n");
		}
		printf("\n");
	}
	return 0;
}
```