#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1), eps = 1e-9;
const int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
const int N = 3e5 + 5, mod = 1e9 + 7, inf = 1e9;
const long long INF = 1e18 + 7;
long long ans;
void calc(string s, char a, long long sum) {
  if ((int)s.size() == 1) {
    if (s[0] != a) sum++;
    ans = min(ans, sum);
    return;
  }
  string left = s.substr(0, (int)s.size() / 2);
  string rigth = s.substr((int)s.size() / 2, (int)s.size() / 2);
  long long l = 0, r = 0;
  for (char c : left)
    if (c != a) l++;
  for (char c : rigth)
    if (c != a) r++;
  calc(left, a + 1, sum + r);
  calc(rigth, a + 1, sum + l);
}
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  ans = INF;
  calc(s, 'a', 0ll);
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) solve();
}

### Prompt

Generate a cpp solution to the following problem:

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;
const double PI = acos(-1), eps = 1e-9;
const int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
const int N = 3e5 + 5, mod = 1e9 + 7, inf = 1e9;
const long long INF = 1e18 + 7;
long long ans;
void calc(string s, char a, long long sum) {
  if ((int)s.size() == 1) {
    if (s[0] != a) sum++;
    ans = min(ans, sum);
    return;
  }
  string left = s.substr(0, (int)s.size() / 2);
  string rigth = s.substr((int)s.size() / 2, (int)s.size() / 2);
  long long l = 0, r = 0;
  for (char c : left)
    if (c != a) l++;
  for (char c : rigth)
    if (c != a) r++;
  calc(left, a + 1, sum + r);
  calc(rigth, a + 1, sum + l);
}
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  ans = INF;
  calc(s, 'a', 0ll);
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) solve();
}

```