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 sys import stdin,stdout,setrecursionlimit
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush,nlargest
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm , accumulate
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
setrecursionlimit(10**9)
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
# from sys import stdin
# input = stdin.readline
#def data(): return sys.stdin.readline().strip()
def data(): return input()
def num():return int(input())
def L(): return list(sp())
def LF(): return list(spf())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def spf(): return map(int, input.readline().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def pmat(A):
    for ele in A:
        print(*ele,end="\n")
def pmat2(A):
    for ele in A:
        for j in ele:
            print(j,end='')
        print()
def iseven(n):
    return n%2==0

def seive(r):
    prime=[1 for i in range(r+1)]
    prime[0]=0
    prime[1]=0
    for i in range(r+1):
        if(prime[i]):
            for j in range(2*i,r+1,i):
                prime[j]=0
    return prime
#solution
#ACPC 
#remeber cut ribbon problem
# set data structures faster than binary search sometimes
#bipartite match dfs
#think in problems with recursive manner.
n,m,x=sp()
ans=''
if x%2==0:
	if n>m:
		for i in range(x//2):
			ans='01'*(x//2)
		ans+='1'*(m-(x//2))
		ans+="0"*(n-(x//2))
	else:
		for i in range(x//2):
			ans='10'*(x//2)
		ans+='0'*(n-(x//2))
		ans+="1"*(m-(x//2))
else:
	if n>m:
		for i in range(x//2):
			ans='01'*(x//2)
		ans+='0'*(n-(x//2))
		ans+="1"*(m-(x//2))
	else:
		for i in range(x//2):
			ans='10'*(x//2)
		ans+='1'*(m-(x//2))
		ans+="0"*(n-(x//2))
print(ans)


endtime = time.time()
#print(f"Runtime of the program is {endtime - starttime}")