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. 
from __future__ import print_function
import sys
import math
import os.path
import random
from copy import deepcopy
from functools import reduce


from collections import Counter, ChainMap, defaultdict
from itertools import cycle, chain
from queue import Queue, PriorityQueue
from heapq import heappush, heappop, heappushpop, heapify, heapreplace, nlargest, nsmallest
import bisect

from statistics import mean, mode, median, median_low, median_high


# CONFIG
sys.setrecursionlimit(10**9)

# LOG 
def log(*args, **kwargs):
    print(*args, file=sys.stderr, **kwargs)


# INPUT
def ni():
    return map(int, input().split())


def nio(offset):
    return map(lambda x: int(x) + offset, input().split())


def nia():
    return list(map(int, input().split()))


# CONVERT
def toString(aList, sep=" "):
    return sep.join(str(x) for x in aList)


def toMapInvertIndex(aList):
    return {k: v for v, k in enumerate(aList)}


# SORT
def sortId(arr):
    return sorted(range(arr), key=lambda k: arr[k])


# MAIN
a,b,x = ni()

s = ""

ch = 1 if a < b else 0

for i in range(x+1):
    s += str(ch)
    ch = 1-ch

c = Counter(s)

# log(s)
need = [a,b]
for i in range(2):
    ch = str(i)
    idi = s.find(ch)
    if idi != -1:
        s = s[:idi] + (ch * (need[i] - c[ch])) + s[idi:]
    # log(ch,idi,s, (ch * (need[i] - c[ch])))

# log(s)
print(s)
