#include <bits/stdc++.h>
using namespace std;
class Solution {
 public:
  int solve(string& S) {
    int n = S.length();
    const int A = 26;
    vector<vector<int>> cnts(n + 1, vector<int>(A, 0));
    for (int i = 0; i < n; ++i) {
      cnts[i + 1] = cnts[i];
      cnts[i + 1][S[i] - 'a'] += 1;
    }
    auto count = [&](int l, int r, int c) {
      if (l > r) {
        return 0;
      }
      return r - l + 1 - (cnts[r + 1][c] - cnts[l][c]);
    };
    map<pair<int, int>, int> M;
    function<int(int, int, int)> search = [&](int l, int len, int ch) {
      if (len == 0) {
        return 0;
      }
      int r = l + len - 1;
      if (len == 1) {
        return count(l, r, ch);
      }
      int m = l + (len >> 1);
      int res = count(l, m - 1, ch) + search(m, len >> 1, ch + 1);
      res = min(res, count(m, r, ch) + search(l, len >> 1, ch + 1));
      return res;
    };
    return search(0, n, 0);
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    string S;
    cin >> S;
    cout << sol.solve(S) << '\n';
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.

The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:

  * The length of s is 1, and it consists of the character c (i.e. s_1=c);
  * The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string; 
  * The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string. 



For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.

In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').

Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.

You have to answer t independent test cases.

Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:

  * the second half of the string ("aaaa") consists of only the character 'a'; 
  * the first half of the string ("cdbb") is 'b'-good string, because: 
    * the second half of the string ("bb") consists of only the character 'b'; 
    * the first half of the string ("cd") is 'c'-good string, because: 
      * the first half of the string ("c") consists of only the character 'c'; 
      * the second half of the string ("d") is 'd'-good string. 

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 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 ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.

Example

Input


6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac


Output


0
7
4
5
1
1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
class Solution {
 public:
  int solve(string& S) {
    int n = S.length();
    const int A = 26;
    vector<vector<int>> cnts(n + 1, vector<int>(A, 0));
    for (int i = 0; i < n; ++i) {
      cnts[i + 1] = cnts[i];
      cnts[i + 1][S[i] - 'a'] += 1;
    }
    auto count = [&](int l, int r, int c) {
      if (l > r) {
        return 0;
      }
      return r - l + 1 - (cnts[r + 1][c] - cnts[l][c]);
    };
    map<pair<int, int>, int> M;
    function<int(int, int, int)> search = [&](int l, int len, int ch) {
      if (len == 0) {
        return 0;
      }
      int r = l + len - 1;
      if (len == 1) {
        return count(l, r, ch);
      }
      int m = l + (len >> 1);
      int res = count(l, m - 1, ch) + search(m, len >> 1, ch + 1);
      res = min(res, count(m, r, ch) + search(l, len >> 1, ch + 1));
      return res;
    };
    return search(0, n, 0);
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    string S;
    cin >> S;
    cout << sol.solve(S) << '\n';
  }
  return 0;
}

```