1352_F. Binary String Reconstruction
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.

You are given three numbers:

  * n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0; 
  * n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1; 
  * n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2. 



For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.

Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.

Input

The first line contains an integer t (1 ≤ 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 n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.

Output

Print t 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


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


Output


1110011110
0011
0110001100101011
10
0000111
1111
000
```python
for testcase in range(int(input())):
    n0, n1, n2 = map(int, input().split())
    if n1 == 0:
        if n0:
            ans = "0" * (n0 + 1)
        else:
            ans = "1" * (n2 + 1)
    else:
        ans = ["0" * (n0 + 1), "1" * (n2 + 1)]
        for i in range(n1 - 1):
            ans.append(str(i & 1))
        ans = "".join(ans)

    print(ans)
```