A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.

You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.

Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.

Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).

Output

For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.

Example

Input


4
5
22222
5
21211
1
2
9
220222021


Output


11111
11111
11000
10211
1
1
110111011
110111010
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, i;
    cin >> n;
    string s, a, b;
    cin >> s;
    for (i = 0; i < n; i++) {
      if (s[i] == '2') {
        a += '0' + 1;
        b += '0' + 1;
      } else if (s[i] == '1') {
        a += '0' + 1;
        b += '0';
        break;
      } else {
        a += '0';
        b += '0';
      }
    }
    if (i != n - 1) {
      for (i = i + 1; i < n; i++) {
        a += '0';
        b += s[i];
      }
    }
    cout << a << "\n" << b << endl;
  }
  return 0;
}
