You are given three integers a, b and x. Your task is to construct a binary string s of length n = a + b such that there are exactly a zeroes, exactly b ones and exactly x indices i (where 1 ≤ i < n) such that s_i ≠ s_{i + 1}. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices i such that 1 ≤ i < n and s_i ≠ s_{i + 1} (i = 1, 2, 3, 4). For the string "111001" there are two such indices i (i = 3, 5).

Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.

Input

The first line of the input contains three integers a, b and x (1 ≤ a, b ≤ 100, 1 ≤ x < a + b).

Output

Print only one string s, where s is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.

Examples

Input

2 2 1


Output

1100


Input

3 3 3


Output

101100


Input

5 3 6


Output

01010100

Note

All possible answers for the first example: 

  * 1100; 
  * 0011. 



All possible answers for the second example: 

  * 110100; 
  * 101100; 
  * 110010; 
  * 100110; 
  * 011001; 
  * 001101; 
  * 010011; 
  * 001011. 
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

import time

(a, b, x) = (int(i) for i in input().split())

start = time.time()

s = []

if a > b:
    j = 0
else:
    j = 1

r = [0, 0]

for i in range(x+1):
    s.append(j)
    r[j] += 1
    j = 1 - j

if b > r[1]:
    i = 0
    while(s[i] == 0):
        i += 1
    s = s[:i]+[1 for j in range(b-r[1])]+s[i:]

if a > r[0]:
    i = 0
    while(s[i] == 1):
        i += 1
    s = s[:i]+[0 for j in range(a-r[0])]+s[i:]

for i in range(a+b):
    print(s[i], end ='')
print()
finish = time.time()
#print(finish - start)
