#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long MOD = 1e9 + 7;
long long size = 10000000;
string s;
long long n;
map<string, long long> mp;
long long dp(long long i, long long k, long long c, bool ch) {
  if (k == 0) {
    if (s[i] == ((char)c))
      return 0;
    else
      return 1;
  }
  if (i >= n) return 0;
  string tem = to_string(i) + '|' + to_string(k) + '|' + to_string(c) + '|' +
               to_string(ch);
  if (mp.find(tem) != mp.end()) return mp[tem];
  long long r = pow(2, k - 1);
  if (ch) {
    long long ans = 0;
    string temp = "";
    for (long long j = i; j < i + r * 2; j++) {
      if (s[j] != ((char)c)) ans++;
      ;
    };
    mp[tem] = ans;
  } else {
    long long v1 = INT_MAX, v2 = INT_MAX;
    ;
    ;
    v1 = dp(i, k - 1, c + 1, 0) + dp(i + r, k - 1, c, 1);
    ;
    ;
    ;
    v2 = dp(i, k - 1, c, 1) + dp(i + r, k - 1, c + 1, 0);
    ;
    mp[tem] = min(v1, v2);
    ;
  }
  return mp[tem];
}
void solve() {
  cin >> n;
  cin >> s;
  long long k = log2(n);
  cout << dp(0, k, 'a', false) << '\n';
  ;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  cin >> T;
  while (T--) {
    solve();
    mp.clear();
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long MOD = 1e9 + 7;
long long size = 10000000;
string s;
long long n;
map<string, long long> mp;
long long dp(long long i, long long k, long long c, bool ch) {
  if (k == 0) {
    if (s[i] == ((char)c))
      return 0;
    else
      return 1;
  }
  if (i >= n) return 0;
  string tem = to_string(i) + '|' + to_string(k) + '|' + to_string(c) + '|' +
               to_string(ch);
  if (mp.find(tem) != mp.end()) return mp[tem];
  long long r = pow(2, k - 1);
  if (ch) {
    long long ans = 0;
    string temp = "";
    for (long long j = i; j < i + r * 2; j++) {
      if (s[j] != ((char)c)) ans++;
      ;
    };
    mp[tem] = ans;
  } else {
    long long v1 = INT_MAX, v2 = INT_MAX;
    ;
    ;
    v1 = dp(i, k - 1, c + 1, 0) + dp(i + r, k - 1, c, 1);
    ;
    ;
    ;
    v2 = dp(i, k - 1, c, 1) + dp(i + r, k - 1, c + 1, 0);
    ;
    mp[tem] = min(v1, v2);
    ;
  }
  return mp[tem];
}
void solve() {
  cin >> n;
  cin >> s;
  long long k = log2(n);
  cout << dp(0, k, 'a', false) << '\n';
  ;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  cin >> T;
  while (T--) {
    solve();
    mp.clear();
  }
  return 0;
}

```