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. 
def findIndexLast(word, l):
	index = -1

	for i in range(len(word)):
		if word[i] == l and i > index:
			index = i

	return index

a, b, x = map(int, raw_input().split())

total = a + b
diff = 0
ans = ""
while diff < x + 1:
	if ans == "" and b >= a:
		ans += "1"
		b -= 1

	elif ans == "" and b <= a:
		ans += "0"
		a -= 1

	else:
		if ans[-1] == "1":
			ans += "0"
			a -= 1

		else:
			ans += "1"
			b -= 1

	diff += 1

if len(ans) < total:
	if a > 0:
		indexLastZero = findIndexLast(ans, "0")
		aux = "0" * a
		ans = ans[:indexLastZero] + aux + ans[indexLastZero:]

	if b > 0:
		indexLastOne = findIndexLast(ans, "1")
		aux = "1" * b
		ans = ans[:indexLastOne] + aux + ans[indexLastOne:]

print ans