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. 
from collections import Counter
import sys
sys.setrecursionlimit(500)

def A():
    N = int(input())
    c = Counter(input().split())
    print(c.most_common(1)[0][1])

visited = set()
solved = 0
sol = ''
cur = ['']
def solve_B(a, b, x):
    global visited
    global solved
    global sol
    global cur
    if (a,b,x,cur[-1]) in visited or solved == 1 or min(a,b,x) < 0 or x>a+b:
        cur.pop()
        return
    visited.add((a,b,x,cur[-1]))
    if a+b+x==0:
        solved = 1
        sol = cur.copy()
        cur.pop()
        return
    cur.append('0')
    solve_B(a-1, b, x if len(cur)==2 or cur[-2] == '0' else x-1)
    cur.append('1')
    solve_B(a, b-1, x if len(cur)==2 or cur[-2] == '1' else x-1)
    cur.pop()
        
def B():
    A, B, X = map(int, input().split())
    solve_B(A,B,X)
    print(''.join(sol))

def C():
    pass

def D():
    pass

def E():
    pass

B()