#P1492D. Genius's Gambit

    ID: 829 Type: RemoteJudge 2000ms 512MiB Tried: 0 Accepted: 0 Difficulty: (None) Uploaded By: Tags>bitmasksconstructive algorithmsgreedymath*1900

Genius's Gambit

No submission language available for this problem.

Description

You are given three integers aa, bb, kk.

Find two binary integers xx and yy (xyx \ge y) such that

  1. both xx and yy consist of aa zeroes and bb ones;
  2. xyx - y (also written in binary form) has exactly kk ones.
You are not allowed to use leading zeros for xx and yy.

The only line contains three integers aa, bb, and kk (0a0 \leq a; 1b1 \leq b; 0ka+b21050 \leq k \leq a + b \leq 2 \cdot 10^5) — the number of zeroes, ones, and the number of ones in the result.

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

Otherwise print "No".

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

Input

The only line contains three integers aa, bb, and kk (0a0 \leq a; 1b1 \leq b; 0ka+b21050 \leq k \leq a + b \leq 2 \cdot 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 xx and yy in base-2.

Otherwise print "No".

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

Samples

Sample Input 1

4 2 3

Sample Output 1

Yes
101000
100001

Sample Input 2

3 2 1

Sample Output 2

Yes
10100
10010

Sample Input 3

3 2 5

Sample Output 3

No

Note

In the first example, x=1010002=25+23=4010x = 101000_2 = 2^5 + 2^3 = 40_{10}, y=1000012=25+20=3310y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40103310=710=22+21+20=111240_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence xyx-y has 33 ones in base-2.

In the second example, x=101002=24+22=2010x = 10100_2 = 2^4 + 2^2 = 20_{10}, y=100102=24+21=18y = 10010_2 = 2^4 + 2^1 = 18, xy=2018=210=102x - 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.