#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define fi(i, a, b) for (int i = a; i < b; i++)
#define fll(i, a, b) for (ll i = a; i < b; i++)
#define vi vector<int>
#define vll vector<ll>
#define srt(v) sort(v.begin(), v.end())
#define rsrt(v) sort(v.begin(), v.end(), greater<int>())
using namespace std;
int main(){
    ios_base::sync_with_stdio(false); 
    cin.tie(0);                       
    cout.tie(0);
    int a, b, k;
    cin >> a >> b >> k;
    vi x(a + b), y(a + b);
    if (k == 0){
        fi(i, 0, b) x[i] = y[i] = 1;
        fi(i, b, a + b) x[i] = y[i] = 0;
        cout << "Yes\n";
        for (int i = 0; i  < a + b; i++) cout << x[i];
        cout << "\n";
        for (int i = 0; i < a + b; i++) cout << x[i];
        return 0;
    }
    if (b == 1){
        if (k != 0){
            cout << "No\n";
            return 0;
        }
    }
    else{
        if (k == a + b || k == a + b - 1){
            cout << "No\n";
            return 0;
        }
        y[a + b - 1] = 1, x[a + b - 1 - k] = 1;
        y[0] = 1, x[0] = 1;
        int p = b - 2;
        fi(i, 1, a + b){
            if (p == 0) break;
            if (x[i] == 0 && y[i] == 0) x[i] = y[i] = 1, p--;
        }
        if (p > 0){
            cout << "No\n";
            return 0;
        }
        cout << "Yes\n";
        fi(i, 0, a + b) cout << x[i];
        cout << "\n";
        fi(i, 0, a + b) cout << y[i];
    }
}
### Prompt

Generate 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
#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define fi(i, a, b) for (int i = a; i < b; i++)
#define fll(i, a, b) for (ll i = a; i < b; i++)
#define vi vector<int>
#define vll vector<ll>
#define srt(v) sort(v.begin(), v.end())
#define rsrt(v) sort(v.begin(), v.end(), greater<int>())
using namespace std;
int main(){
    ios_base::sync_with_stdio(false); 
    cin.tie(0);                       
    cout.tie(0);
    int a, b, k;
    cin >> a >> b >> k;
    vi x(a + b), y(a + b);
    if (k == 0){
        fi(i, 0, b) x[i] = y[i] = 1;
        fi(i, b, a + b) x[i] = y[i] = 0;
        cout << "Yes\n";
        for (int i = 0; i  < a + b; i++) cout << x[i];
        cout << "\n";
        for (int i = 0; i < a + b; i++) cout << x[i];
        return 0;
    }
    if (b == 1){
        if (k != 0){
            cout << "No\n";
            return 0;
        }
    }
    else{
        if (k == a + b || k == a + b - 1){
            cout << "No\n";
            return 0;
        }
        y[a + b - 1] = 1, x[a + b - 1 - k] = 1;
        y[0] = 1, x[0] = 1;
        int p = b - 2;
        fi(i, 1, a + b){
            if (p == 0) break;
            if (x[i] == 0 && y[i] == 0) x[i] = y[i] = 1, p--;
        }
        if (p > 0){
            cout << "No\n";
            return 0;
        }
        cout << "Yes\n";
        fi(i, 0, a + b) cout << x[i];
        cout << "\n";
        fi(i, 0, a + b) cout << y[i];
    }
}
```