#include <bits/stdc++.h>
using namespace std;
struct XX {
  int x;
  int xx;
  int f;
  int i;
};
class xxGreater {
 public:
  bool operator()(const XX& riLeft, const XX& riRight) const {
    return (riLeft.x) < (riRight.x);
  }
};
class xxGreater2 {
 public:
  bool operator()(const XX& riLeft, const XX& riRight) const {
    return (riLeft.i) < (riRight.i);
  }
};
int main(int argc, const char* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int a[200000];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    string s;
    cin >> s;
    XX x[200000];
    for (int i = 0; i < n; i++) {
      if (s[i] == '1') {
        x[i].x = a[i];
        x[i].i = i;
        x[i].f = 1;
      } else {
        x[i].x = a[i];
        x[i].i = i;
        x[i].f = 0;
      }
    }
    sort(x, x + n, xxGreater());
    int ind = 1;
    for (int i = 0; i < n; i++) {
      if (x[i].f == 0) {
        x[i].xx = ind;
        ind++;
      }
    }
    for (int i = 0; i < n; i++) {
      if (x[i].f == 1) {
        x[i].xx = ind;
        ind++;
      }
    }
    sort(x, x + n, xxGreater2());
    for (int i = 0; i < n; i++) {
      cout << x[i].xx << " ";
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

Berland Music is a music streaming service built specifically to support Berland local artist. Its developers are currently working on a song recommendation module.

So imagine Monocarp got recommended n songs, numbered from 1 to n. The i-th song had its predicted rating equal to p_i, where 1 ≤ p_i ≤ n and every integer from 1 to n appears exactly once. In other words, p is a permutation.

After listening to each of them, Monocarp pressed either a like or a dislike button. Let his vote sequence be represented with a string s, such that s_i=0 means that he disliked the i-th song, and s_i=1 means that he liked it.

Now the service has to re-evaluate the song ratings in such a way that:

  * the new ratings q_1, q_2, ..., q_n still form a permutation (1 ≤ q_i ≤ n; each integer from 1 to n appears exactly once); 
  * every song that Monocarp liked should have a greater rating than every song that Monocarp disliked (formally, for all i, j such that s_i=1 and s_j=0, q_i>q_j should hold). 



Among all valid permutations q find the one that has the smallest value of ∑_{i=1}^n |p_i-q_i|, where |x| is an absolute value of x.

Print the permutation q_1, q_2, ..., q_n. If there are multiple answers, you can print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.

The first line of each testcase contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of songs.

The second line of each testcase contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n) — the permutation of the predicted ratings.

The third line contains a single string s, consisting of n characters. Each character is either a 0 or a 1. 0 means that Monocarp disliked the song, and 1 means that he liked it.

The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.

Output

For each testcase, print a permutation q — the re-evaluated ratings of the songs. If there are multiple answers such that ∑_{i=1}^n |p_i-q_i| is minimum possible, you can print any of them.

Example

Input


3
2
1 2
10
3
3 1 2
111
8
2 3 1 8 5 4 7 6
01110001


Output


2 1
3 1 2
1 6 5 8 3 2 4 7

Note

In the first testcase, there exists only one permutation q such that each liked song is rating higher than each disliked song: song 1 gets rating 2 and song 2 gets rating 1. ∑_{i=1}^n |p_i-q_i|=|1-2|+|2-1|=2.

In the second testcase, Monocarp liked all songs, so all permutations could work. The permutation with the minimum sum of absolute differences is the permutation equal to p. Its cost is 0.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct XX {
  int x;
  int xx;
  int f;
  int i;
};
class xxGreater {
 public:
  bool operator()(const XX& riLeft, const XX& riRight) const {
    return (riLeft.x) < (riRight.x);
  }
};
class xxGreater2 {
 public:
  bool operator()(const XX& riLeft, const XX& riRight) const {
    return (riLeft.i) < (riRight.i);
  }
};
int main(int argc, const char* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int a[200000];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    string s;
    cin >> s;
    XX x[200000];
    for (int i = 0; i < n; i++) {
      if (s[i] == '1') {
        x[i].x = a[i];
        x[i].i = i;
        x[i].f = 1;
      } else {
        x[i].x = a[i];
        x[i].i = i;
        x[i].f = 0;
      }
    }
    sort(x, x + n, xxGreater());
    int ind = 1;
    for (int i = 0; i < n; i++) {
      if (x[i].f == 0) {
        x[i].xx = ind;
        ind++;
      }
    }
    for (int i = 0; i < n; i++) {
      if (x[i].f == 1) {
        x[i].xx = ind;
        ind++;
      }
    }
    sort(x, x + n, xxGreater2());
    for (int i = 0; i < n; i++) {
      cout << x[i].xx << " ";
    }
    cout << endl;
  }
  return 0;
}

```