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. 
'''input
3 3 3
'''
import sys
from collections import defaultdict as dd
from itertools import  permutations as pp
from itertools import combinations as cc
from collections import Counter as ccd
from random import randint as rd
from bisect import bisect_left as bl
import heapq
mod=10**9+7

def ri(flag=0):
	if flag==0:
		return [int(i) for i in input().split()]
	else:
		return int(input())

a,b,k=ri()

ans=[]
ones=0
zeroes=0
prev=0
for i in range(k):
	ones+=(prev==1)
	zeroes+=(prev==0)
	ans.append(prev)
	prev=prev^1

if len(ans)!=0 and ans[-1]==1:
	for i in range(b-ones):
		ans.append(1)
	for i in range(a-zeroes):
		ans.append(0)
else:
	for i in range(a-zeroes):
		ans.append(0)
	for i in range(b-ones):
		ans.append(1)

ans1=[]
ones=0
zeroes=0
prev=1
for i in range(k):
	ones+=(prev==1)
	zeroes+=(prev==0)
	ans1.append(prev)
	prev=prev^1

if len(ans1)!=0 and ans1[-1]==1:
	for i in range(b-ones):
		ans1.append(1)
	for i in range(a-zeroes):
		ans1.append(0)
else:
	for i in range(a-zeroes):
		ans1.append(0)
	for i in range(b-ones):
		ans1.append(1)

def check(ar):
	#print(ar)
	temp=0
	for i in range(1,len(ar)):
		if ar[i]!=ar[i-1]:
			temp+=1
	if ar.count(0)==a and ar.count(1)==b and temp==k:
		return 1
	else:
		return 0
if check(ans):
	print("".join([str(i) for i in ans]))
elif check(ans1) :
	print("".join([str(i) for i in ans1]))
else:
	print(1/0)