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. 
import math as ma
from sys import exit
from decimal import Decimal as dec


def li():
	return list(map(int , input().split()))


def num():
	return map(int , input().split())


def nu():
	return int(input())

a,b,x=num()
pp=0
ut=0
zulmi=0
if(a>=b):
	pp=0
	ut=1
	zulmi=b
else:
	pp=1
	ut=0
	zulmi=a
v=[ut]*(a+b)
if(x%2==1):
	z=(x-1)//2
else:
	z=x//2
if(x%2==0):
	v[0]=pp
	z=z-1
	p=z
	ll=2
	while(p>0):
		v[ll]=pp
		p-=1
		ll+=2
	#print(v)
	for j in range(zulmi+z+1,(a+b)):
		#print(j)
		v[j]=pp
else:
	ll=1
	p=z
	#print(p)
	while (p > 0):
		v[ll] = pp
		p -= 1
		ll += 2
	for j in range(zulmi+z, (a + b)):
		v[j] = pp
print(*v,sep="")