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. 
l=[int(x) for x in input().split()]
a=l[0]
b=l[1]
n=a+b
y=l[2]
x=y
res=""
t=""
	
if(x!=1):
	while(x>0):
		if a>b:
			t+='0'
			x-=1
			a-=1
			if(x>0):
				t+='1'
				b-=1				
				x-=1
		else:
			t+='1'
			b-=1
			x-=1
			if(x>0):
				t+='0'
				a-=1	
				x-=1

# print("t",t)
g=0
if(x==1):
	for i in range(b):
		t+='1'
	for i in range(a):
		t += '0'	
elif(y%2==1):
	if(t[-1]=='0'):
		for i in range(b):
			t+='1'
		for i in range(a):
			t = '0'+t
	else:
		for i in range(a):
			t+='0'
		for i in range(b):
			# print("yes")
			t = '1'+ t 
else:
	# print("what")
	g=1
	if(t[-1]=='0'):
		# print("in 0")
		for i in range(b):
			t+='1'
		s=""
		for i in range(a):
			s+='0'
		t = '1' + s + t[1:]	
	else:
		# print("tes")
		for i in range(a):
			t+='0'
		s=""
		for i in range(b):
			s+='1'		
		# print("t1",t[1:])
		mine = t[1:]
		t = '0' + s + mine

print(t)

# cnt=0

# for i in range(n):
# 	if (i<n-1 and t[i]!=t[i+1]):
# 		cnt+=1

# print(cnt)
