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. 
n,m,k = input().split()
a = (int)(n)
b = (int)(m)
x = (int)(k)
ls3 = []
ls = []
ls1 = []
ls1.append(a)
ls1.append(b)
ls2 = []
ls2.append(a)
ls2.append(b)
l = a+b 
ls.append(0)
ls1[0] -= 1
ls3.append(1)
ls2[1] -= 1
count = 0 

i = 1
while i < l:
    if count == x-1:
        if ls3[i-1] == 0:
            f = 0
            q = 1
        else:
            f = 1
            q = 0
        while ls2[f] > 0:
            ls3.append(f)
            ls2[f] -= 1
        while ls2[q] > 0:
            ls3.append(q)
            ls2[q] -= 1
        break
    else:
        if ls3[i-1] == 1:
            ls3.append(0)
            ls2[0] -= 1
        else:
            ls3.append(1)
            ls2[1] -= 1
        count += 1
        i += 1
count = 0
i = 1
while i < l:
    if count == x-1:
        if ls[i-1] == 0:
            f = 0
            q = 1
        else:
            f = 1
            q = 0
        while ls1[f] > 0:
            ls.append(f)
            ls1[f] -= 1
        while ls1[q] > 0:
            ls.append(q)
            ls1[q] -= 1
        break
    else:
        if ls[i-1] == 1:
            ls.append(0)
            ls1[0] -= 1
        else:
            ls.append(1)
            ls1[1] -= 1
        count += 1
        i += 1
    # print(ls)
    # print(count)
count = 0 
# print(ls)
# print(ls3)
for j in range(1,l):
    if ls[j] != ls[j-1]:
        count += 1
if count == x:
    for j in range(0,l):
        print(ls[j],end='')
else:
    for j in range(0,l):
        print(ls3[j],end='')