#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p);
template <class T>
ostream &operator<<(ostream &os, vector<T> &v);
template <class T>
ostream &operator<<(ostream &os, multiset<T> &v);
template <class T1, class T2>
ostream &operator<<(ostream &os, map<T1, T2> &v);
const long long mod = 1e9 + 7;
const int N = 131072 + 5;
int dp[N][19];
int n;
string s;
int koto(int l, int r, int now) {
  int cnt = 0;
  for (int i = l; i <= r; i++) {
    assert(i < n);
    if (s[i] - 'a' != now) cnt++;
  }
  return cnt;
}
int solve(int l, int step) {
  int r = l + n / (1 << step) - 1;
  int now = step;
  if (l > r) {
    assert(false);
  }
  if (l == r) {
    if (s[l] - 'a' == now)
      return 0;
    else {
      return 1;
    }
  }
  int &ret = dp[l][step];
  if (ret != -1) return ret;
  int mid = (l + r) / 2;
  int ans = min(koto(l, mid, now) + solve(mid + 1, step + 1),
                koto(mid + 1, r, now) + solve(l, step + 1));
  return ret = ans;
}
void TEST_CASES(int cas) {
  scanf("%d", &n);
  cin >> s;
  for (int i = 0; i <= n; i++) memset(dp[i], -1, sizeof dp[i]);
  cout << solve(0, 0) << "\n";
}
int main() {
  int t = 1, cas = 0;
  scanf("%d", &t);
  while (t--) {
    TEST_CASES(++cas);
  }
  return 0;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << "{" << p.first << ", " << p.second << "} ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) {
    os << v[i] << " ";
  }
  os << " ]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, multiset<T> &v) {
  os << "[ ";
  for (T i : v) {
    os << i << " ";
  }
  os << " ]";
  return os;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, map<T1, T2> &v) {
  for (auto i : v) {
    os << "Key : " << i.first << " , Value : " << i.second << endl;
  }
  return os;
}

### Prompt

Please create a solution in CPP 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;
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p);
template <class T>
ostream &operator<<(ostream &os, vector<T> &v);
template <class T>
ostream &operator<<(ostream &os, multiset<T> &v);
template <class T1, class T2>
ostream &operator<<(ostream &os, map<T1, T2> &v);
const long long mod = 1e9 + 7;
const int N = 131072 + 5;
int dp[N][19];
int n;
string s;
int koto(int l, int r, int now) {
  int cnt = 0;
  for (int i = l; i <= r; i++) {
    assert(i < n);
    if (s[i] - 'a' != now) cnt++;
  }
  return cnt;
}
int solve(int l, int step) {
  int r = l + n / (1 << step) - 1;
  int now = step;
  if (l > r) {
    assert(false);
  }
  if (l == r) {
    if (s[l] - 'a' == now)
      return 0;
    else {
      return 1;
    }
  }
  int &ret = dp[l][step];
  if (ret != -1) return ret;
  int mid = (l + r) / 2;
  int ans = min(koto(l, mid, now) + solve(mid + 1, step + 1),
                koto(mid + 1, r, now) + solve(l, step + 1));
  return ret = ans;
}
void TEST_CASES(int cas) {
  scanf("%d", &n);
  cin >> s;
  for (int i = 0; i <= n; i++) memset(dp[i], -1, sizeof dp[i]);
  cout << solve(0, 0) << "\n";
}
int main() {
  int t = 1, cas = 0;
  scanf("%d", &t);
  while (t--) {
    TEST_CASES(++cas);
  }
  return 0;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << "{" << p.first << ", " << p.second << "} ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) {
    os << v[i] << " ";
  }
  os << " ]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, multiset<T> &v) {
  os << "[ ";
  for (T i : v) {
    os << i << " ";
  }
  os << " ]";
  return os;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, map<T1, T2> &v) {
  for (auto i : v) {
    os << "Key : " << i.first << " , Value : " << i.second << endl;
  }
  return os;
}

```