#include <bits/stdc++.h>
using namespace std;
void start() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
template <class T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto& i : a) in >> i;
  return in;
}
template <class T>
ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto& i : a) out << i << ' ';
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, vector<pair<T, U> >& a) {
  for (auto& i : a) in >> i.first >> i.second;
  return in;
}
template <class T, class U>
ostream& operator<<(ostream& out, vector<pair<T, U> >& a) {
  for (auto& i : a) out << i.first << ' ' << i.second << "\n";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, pair<T, U>& a) {
  out << a.first << ' ' << a.second << "\n";
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, pair<T, U>& a) {
  in >> a.first >> a.second;
  return in;
}
long long n;
string second;
long long lg = 0;
long long rec(long long l, long long r, long long cur) {
  if (l + 1 == r) {
    if (cur + 'a' != second[l]) {
      return 1;
    }
    return 0;
  }
  map<char, long long> cnt1;
  map<char, long long> cnt2;
  for (long long i = l; i < (l + r) / 2; i++) {
    cnt1[second[i]]++;
  }
  for (long long i = (l + r) / 2; i < r; i++) {
    cnt2[second[i]]++;
  }
  long long ans1 = rec(l, (l + r) / 2, cur + 1);
  long long ans2 = rec((l + r) / 2, r, cur + 1);
  long long val = (r - l) / 2;
  return min(ans1 + (val - cnt2['a' + cur]), ans2 + (val - cnt1['a' + cur]));
}
void solve() {
  cin >> n;
  cin >> second;
  while ((1 << lg) != n) {
    lg++;
  }
  long long ans = rec(0, n, 0);
  cout << ans << "\n";
}
signed main() {
  start();
  long long q = 1;
  cin >> q;
  while (q--) {
    solve();
  }
}

### Prompt

In cpp, your task is to solve 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;
void start() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
template <class T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto& i : a) in >> i;
  return in;
}
template <class T>
ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto& i : a) out << i << ' ';
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, vector<pair<T, U> >& a) {
  for (auto& i : a) in >> i.first >> i.second;
  return in;
}
template <class T, class U>
ostream& operator<<(ostream& out, vector<pair<T, U> >& a) {
  for (auto& i : a) out << i.first << ' ' << i.second << "\n";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, pair<T, U>& a) {
  out << a.first << ' ' << a.second << "\n";
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, pair<T, U>& a) {
  in >> a.first >> a.second;
  return in;
}
long long n;
string second;
long long lg = 0;
long long rec(long long l, long long r, long long cur) {
  if (l + 1 == r) {
    if (cur + 'a' != second[l]) {
      return 1;
    }
    return 0;
  }
  map<char, long long> cnt1;
  map<char, long long> cnt2;
  for (long long i = l; i < (l + r) / 2; i++) {
    cnt1[second[i]]++;
  }
  for (long long i = (l + r) / 2; i < r; i++) {
    cnt2[second[i]]++;
  }
  long long ans1 = rec(l, (l + r) / 2, cur + 1);
  long long ans2 = rec((l + r) / 2, r, cur + 1);
  long long val = (r - l) / 2;
  return min(ans1 + (val - cnt2['a' + cur]), ans2 + (val - cnt1['a' + cur]));
}
void solve() {
  cin >> n;
  cin >> second;
  while ((1 << lg) != n) {
    lg++;
  }
  long long ans = rec(0, n, 0);
  cout << ans << "\n";
}
signed main() {
  start();
  long long q = 1;
  cin >> q;
  while (q--) {
    solve();
  }
}

```