#include <bits/stdc++.h>
#define _for(i, a) for (int i = 0, lennn = (a); i < lennn; ++i)
#define _rep(i, a, b) for (int i = (a), lennn = (b); i <= lennn; ++i)
using namespace std;
typedef long long LL;

inline LL read() {
    LL x(0), f(1); char ch(getchar());
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

int a, b, k;
string ans, ant;

void init() {
    ans.clear();
    ant.clear();
}

void sol() {
    init();
    if(k == a + b) {
        cout << "No\n";
        return;
    }
    int l = min(a + b - k - 1, k ? b - 1 : b);
    _for(i, l) ans.push_back('1');
    _for(i, a) ans.push_back('0');
    _for(i, b - l) ans.push_back('1');
    _for(i, a + b) ant.push_back('0');
    int f = 0;
    for(int i = ans.size() - 1; i >= 0; --i) {
        int tem = ans[i] - '0' + (ans.size() - i <= k) + f;
        if(tem == 1) ant[i] = '1', f = 0;
        else if(tem == 2) f = 1;
        else if(tem == 3) ant[i] = '1', f = 1;
    }
    if(f || ans[0] == '0' || ant[0] == '0') {
        cout << "No\n";
        return;
    }
    cout << "Yes\n";
    cout << ant << "\n" << ans << "\n";
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    while(cin>>a>>b>>k) {
        sol();
    }
    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
#include <bits/stdc++.h>
#define _for(i, a) for (int i = 0, lennn = (a); i < lennn; ++i)
#define _rep(i, a, b) for (int i = (a), lennn = (b); i <= lennn; ++i)
using namespace std;
typedef long long LL;

inline LL read() {
    LL x(0), f(1); char ch(getchar());
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

int a, b, k;
string ans, ant;

void init() {
    ans.clear();
    ant.clear();
}

void sol() {
    init();
    if(k == a + b) {
        cout << "No\n";
        return;
    }
    int l = min(a + b - k - 1, k ? b - 1 : b);
    _for(i, l) ans.push_back('1');
    _for(i, a) ans.push_back('0');
    _for(i, b - l) ans.push_back('1');
    _for(i, a + b) ant.push_back('0');
    int f = 0;
    for(int i = ans.size() - 1; i >= 0; --i) {
        int tem = ans[i] - '0' + (ans.size() - i <= k) + f;
        if(tem == 1) ant[i] = '1', f = 0;
        else if(tem == 2) f = 1;
        else if(tem == 3) ant[i] = '1', f = 1;
    }
    if(f || ans[0] == '0' || ant[0] == '0') {
        cout << "No\n";
        return;
    }
    cout << "Yes\n";
    cout << ant << "\n" << ans << "\n";
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    while(cin>>a>>b>>k) {
        sol();
    }
    return 0;
}

```