#include <bits/stdc++.h>
using namespace std;
void input(vector<long long> &v) {
  for (auto &i : v) cin >> i;
}
void output(vector<long long> &v) {
  for (auto i : v) cout << i << ' ';
  cout << endl;
}
long long t;
string s;
long long f(char c, long long l, long long r, long long ans,
            vector<vector<long long> > &v, long long cnt) {
  if (l == r) return ans + (c != s[l]);
  long long mid = (l + r) / 2;
  long long lc = v[mid][c - 'a'] - (l != 0 ? v[l - 1][c - 'a'] : 0),
            rc = v[r][c - 'a'] - v[mid][c - 'a'];
  return min(f(c + 1, l, mid, ans + cnt - rc, v, cnt >> 1),
             f(c + 1, mid + 1, r, ans + cnt - lc, v, cnt >> 1));
}
void solve() {
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cin >> s;
    vector<vector<long long> > v(n, vector<long long>(26));
    for (long long i = 0; i <= n - 1; i++) {
      if (i) {
        for (long long j = 0; j <= 25; j++) v[i][j] = v[i - 1][j];
        ++v[i][s[i] - 'a'];
      } else
        v[i][s[i] - 'a'] = 1;
    }
    cout << f('a', 0, n - 1, 0, v, n >> 1) << '\n';
  }
}
int main() {
  ios::sync_with_stdio(false);
  solve();
}

### 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;
void input(vector<long long> &v) {
  for (auto &i : v) cin >> i;
}
void output(vector<long long> &v) {
  for (auto i : v) cout << i << ' ';
  cout << endl;
}
long long t;
string s;
long long f(char c, long long l, long long r, long long ans,
            vector<vector<long long> > &v, long long cnt) {
  if (l == r) return ans + (c != s[l]);
  long long mid = (l + r) / 2;
  long long lc = v[mid][c - 'a'] - (l != 0 ? v[l - 1][c - 'a'] : 0),
            rc = v[r][c - 'a'] - v[mid][c - 'a'];
  return min(f(c + 1, l, mid, ans + cnt - rc, v, cnt >> 1),
             f(c + 1, mid + 1, r, ans + cnt - lc, v, cnt >> 1));
}
void solve() {
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cin >> s;
    vector<vector<long long> > v(n, vector<long long>(26));
    for (long long i = 0; i <= n - 1; i++) {
      if (i) {
        for (long long j = 0; j <= 25; j++) v[i][j] = v[i - 1][j];
        ++v[i][s[i] - 'a'];
      } else
        v[i][s[i] - 'a'] = 1;
    }
    cout << f('a', 0, n - 1, 0, v, n >> 1) << '\n';
  }
}
int main() {
  ios::sync_with_stdio(false);
  solve();
}

```