F. Binary String Reconstruction

For some binary string ss (i.e. each character sisi is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 22 were written. For each pair (substring of length 22), the number of '1' (ones) in it was calculated.

You are given three numbers:

• n0n0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 00;
• n1n1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 11;
• n2n2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 22.

For example, for the string s=s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n0=1n0=1, n1=3n1=3, n2=5n2=5.

Your task is to restore any suitable binary string ss from the given values n0,n1,n2n0,n1,n2. It is guaranteed that at least one of the numbers n0,n1,n2n0,n1,n2 is greater than 00. Also, it is guaranteed that a solution exists.

Input

The first line contains an integer tt (1≤t≤10001≤t≤1000) — the number of test cases in the input. Then test cases follow.

Each test case consists of one line which contains three integers n0,n1,n2n0,n1,n2 (0≤n0,n1,n2≤1000≤n0,n1,n2≤100; n0+n1+n2>0n0+n1+n2>0). It is guaranteed that the answer for given n0,n1,n2n0,n1,n2 exists.

Output

Print tt lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.

Example

input

Copy

7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0

output

Copy

1110011110
0011
0110001100101011
10
0000111
1111
000

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
//#include <xfunctional>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 998244353;
const int N = 2e5+5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}

ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m%n);
}
ll lcm(ll m, ll n)
{
    return m*n / gcd(m, n);
}

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        string s="";
        int a, b, c;
        cin >> a >> b >> c;
        if (a > 0)
        {
            rep(i, 0, a)
            {
                s += '0';
            }
        }
        if (c > 0)
        {
            rep(i, 0, c)
            {
                s += '1';
            }
        }
        int fg;
        if (a==0 && c==0)
        {
            b++;
            fg = 1;
        }
        if(a>0 && c>0)
            b--,fg = 1;
        if (a == 0 && c > 0)
            fg = 1;
        if (a > 0 && c == 0)
            fg = 0;
        rep(i, 1, b)
        {
            if (fg)
            {
                s += '0';
                fg = 0;
            }
            else
            {
                s += '1';
                fg = 1;
            }
        }
        cout << s << endl;
    }
    return 0;
}