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 si≠si+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 si≠si+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.

【代码】：

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,x,n) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long ll;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const ll LNF = 1e18;

const int N = 1e3 + 20;

const int maxm = 1e6 + 10;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int dx[] = {-1,1,0,0,1,1,-1,-1};

const int dy[] = {0,0,1,-1,1,-1,1,-1};

int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

/*

对于x，我们很容易就可以想到先输出x/2对0和1（n对0和1交错出现可以提供2*n-1个符合题意的ai）

然后将剩余的0和剩余的1连续输出（提供1个符合条件的ai，这样就刚好是x个符合条件的ai了，

需要注意的是，我们要优先将个数多的放在前面，

例如，有10个1,5个0的话，我们先输出x/2个“10”，否则，输出x/2个“01”）。

*/

int main()

{

	int a,b,x;

    while(cin>>a>>b>>x)

    {

        if(x%2==0)//5 2 4

        {

            if(a>b)

            {

                for(int i=0;i<x/2;i++)

                    cout<<"01";

                cout<<string(b-x/2,'1');

                cout<<string(a-x/2,'0');

            }

            else{

                for(int i=0;i<x/2;i++)

                    cout<<"10";

                cout<<string(a-x/2,'0');

                cout<<string(b-x/2,'1');

            }

        }

        else

        {

            if(a>b)//5 2 3

            {

                for(int i=0;i<x/2;i++)

                    cout<<"01";

                cout<<string(a-x/2,'0');

                cout<<string(b-x/2,'1');

            }

            else

            {

                for(int i=0;i<x/2;i++)

                    cout<<"10";

                cout<<string(b-x/2,'1');

                cout<<string(a-x/2,'0');

            }

        }

        cout<<endl;

    }

}