//jai baba loknath
#include<bits/stdc++.h>
using namespace std;
typedef       long long int      ll;
#define       optimize           ios::sync_with_stdio(0);cin.tie(0);
#define       Yes                cout<<"Yes"<<endl;
#define       YES                cout<<"YES"<<endl;
#define       No                 cout<<"No"<<endl;
#define       NO                 cout<<"NO"<<endl;
int main(void)
{
   optimize
   ll t,n,m,k,a,b;
   string x,y;
   {
      cin>>a>>b>>k;
      if(k==0)
      {
         for(ll i=1;i<=b;i++)
            x+='1';
         for(ll i=1;i<=a;i++)
            x+='0';
         Yes
         cout<<x<<endl;
         cout<<x<<endl;
         return 0;
      }
      if(k+2>a+b || a<1 || b<2)
      {
         No
         return 0;
      }
      b-=2;
      a-=1;
      x="11";
      y="10";
      for(ll i=1;i<=k-1;i++)
      {
         //cout<<"in"<<endl;
         if(a>0)
         {
            x+='0';
            y+='0';
            a--;
         }
         else
         {
            x+='1';
            y+='1';
            b--;
         }
      }
      //cout<<a<<" "<<b<<endl;
      //cout<<x<<endl<<y<<endl;
      x+='0';
      y+='1';
      //cout<<x<<endl<<y<<endl;
      while(a>0)
      {
         x+='0';
         y+='0';
         a--;
      }
      while(b>0)
      {
         x+='1';
         y+='1';
         b--;
      }
      Yes
      cout<<x<<endl;
      cout<<y<<endl;
   }
   return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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
//jai baba loknath
#include<bits/stdc++.h>
using namespace std;
typedef       long long int      ll;
#define       optimize           ios::sync_with_stdio(0);cin.tie(0);
#define       Yes                cout<<"Yes"<<endl;
#define       YES                cout<<"YES"<<endl;
#define       No                 cout<<"No"<<endl;
#define       NO                 cout<<"NO"<<endl;
int main(void)
{
   optimize
   ll t,n,m,k,a,b;
   string x,y;
   {
      cin>>a>>b>>k;
      if(k==0)
      {
         for(ll i=1;i<=b;i++)
            x+='1';
         for(ll i=1;i<=a;i++)
            x+='0';
         Yes
         cout<<x<<endl;
         cout<<x<<endl;
         return 0;
      }
      if(k+2>a+b || a<1 || b<2)
      {
         No
         return 0;
      }
      b-=2;
      a-=1;
      x="11";
      y="10";
      for(ll i=1;i<=k-1;i++)
      {
         //cout<<"in"<<endl;
         if(a>0)
         {
            x+='0';
            y+='0';
            a--;
         }
         else
         {
            x+='1';
            y+='1';
            b--;
         }
      }
      //cout<<a<<" "<<b<<endl;
      //cout<<x<<endl<<y<<endl;
      x+='0';
      y+='1';
      //cout<<x<<endl<<y<<endl;
      while(a>0)
      {
         x+='0';
         y+='0';
         a--;
      }
      while(b>0)
      {
         x+='1';
         y+='1';
         b--;
      }
      Yes
      cout<<x<<endl;
      cout<<y<<endl;
   }
   return 0;
}

```