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. 
a,b,x=list(map(int,input().split()))
if x==1:
	for i in range(a):
		print("0",end="")
	for j in range(b):
		print("1",end="")
	print()
else:
	one=(x//2)+1
	zero=x-one+1
	if one>b or zero>a:
		zero=(x//2)+1
		one=x-zero+1
	f,s=0,0
	if zero>one:
		f=1
	else:
		s=1
	grp0=[0]*zero
	grp1=[0]*one
	z=a//zero
	for i in range(zero-1):
		grp0[i]=z
	grp0[zero-1]=a-z*(zero-1)
	o=b//one
	for i in range(one-1):
		grp1[i]=o
	grp1[one-1]=b-o*(one-1)
	while(zero and one):
		if f==1:
			for k in range(grp0[zero-1]):
				print("0",end="")
			s=1
			f=0
			zero-=1
		else:
			for k in range(grp1[one-1]):
				print("1",end="")
			s=0
			f=1
			one-=1
	while(zero):
		for k in range(grp0[zero-1]):
			print("0",end="")
		zero-=1
	while(one):
		for k in range(grp1[one-1]):
			print("1",end="")
		one-=1
	print()


