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>
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
using namespace std;
using ll = long long;
using P = pair<int, int>;
using Pl = pair<long long, long long>;
using veci = vector<int>;
using vecl = vector<long long>;
using vecveci = vector<vector<int>>;
using vecvecl = vector<vector<long long>>;
const int MOD = 1000000009;
const double pi = acos(-1);
const ll INF = 1e9;
void solve() {
  int n;
  cin >> n;
  string x;
  cin >> x;
  string a = "1", b = "1";
  for (int i = 0; i < (n); ++i) {
    if (i == 0) continue;
    if (x[i] == '0') {
      a += '0', b += '0';
    } else if (x[i] == '1') {
      if (a > b)
        a += '0', b += '1';
      else
        a += '1', b += '0';
    } else {
      if (a == b)
        a += '1', b += '1';
      else if (a > b)
        a += '0', b += '2';
      else
        a += '2', b += '0';
    }
  }
  cout << a << endl;
  cout << b << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
}
