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. 
yoar=list(map(int,input().split()))
a=yoar[0]
b=yoar[1]
x=yoar[2]
x1=x
# if (x%2==0):
# 	y=x//2
# 	left1=a-y
# 	left0=b-y
# 	print("1"*left1+"01"*y+"0"*left0)
# else:	
# 	y=x//2
# 	y1=y+1
# 	y0=y
# 	left1=a-y1
# 	left0=a-y0
# 	print("1"*left1+"10"*y+"1"+"0"*left0)

st= "1"
a1=a #0
b1=b-1 #1
for i in range(a+b-1):
	if x>1:
		if st[-1]=="1":
			st=st+"0"
			a1-=1
		else:
			st=st+"1"	
			b1-=1
		x-=1	

	else:
		if st[-1]=="0" and a1>0:
			st=st+"0"
			a1-=1
		elif st[-1]=="1" and b1>0:
			st=st+"1"
			b1-=1
		elif a1>0:
			st=st+"0"
			a1-=1
		else:
			st=st+"1"					
			b1-=1
count=0
for i in range(1,a+b):
	if (st[i]!=st[i-1])	:		
		count+=1
if (count==x1):
	print(st)
else:
	print(st[-1]+st[:-1])			