#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class T>
ostream &operator<<(ostream &os, const unordered_set<T> &S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, const pair<L, R> &P) {
  return os << "(" << P.first << "," << P.second << ") ";
}
template <class L, class R>
ostream &operator<<(ostream &os, const map<L, R> &M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
template <class T>
ostream &operator<<(ostream &os, stack<T> S) {
  while (!S.empty()) {
    os << S.top() << "\n";
    S.pop();
  }
  return os;
}
template <class T>
ostream &operator<<(ostream &os, queue<T> Q) {
  while (!Q.empty()) {
    os << Q.front() << " ";
    Q.pop();
  }
  return os;
}
const long long mod = 1e9 + 7;
inline long long modulo(long long x) {
  x = x % mod;
  return x < 0 ? x + mod : x;
}
inline long long add(long long x, long long y) {
  x += y;
  return modulo(x);
}
inline long long sub(long long x, long long y) {
  x -= y;
  return modulo(x);
}
inline long long mul(long long x, long long y) {
  x = x * 1ll * y;
  return modulo(x);
}
inline long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long temp = power(x, y / 2);
  temp = mul(temp, temp);
  if (y % 2 == 1) temp = mul(temp, x);
  return temp;
}
inline long long inv(long long x) { return power(x, mod - 2); }
long long makeall(vector<char> &cache, long long i, long long j, char c) {
  long long count = 0;
  for (long long k = i; i < j ? k <= j : k >= j; i < j ? k++ : k--) {
    if (cache[k] != c) count++;
  }
  return count;
}
long long make(vector<char> &cache, long long i, long long j, char c) {
  if (i == j) return (cache[i] == c) ? 0 : 1;
  long long mid = (i + j) / 2;
  long long temp = makeall(cache, i, mid, c) + make(cache, mid + 1, j, c + 1);
  temp = min(makeall(cache, mid + 1, j, c) + make(cache, i, mid, c + 1), temp);
  return temp;
}
void solve() {
  long long n;
  cin >> n;
  vector<char> cache(n, '@');
  for (long long i = 0; i < n; i++) cin >> cache[i];
  cout << make(cache, 0, n - 1, 'a') << endl;
}
int32_t main() {
  long long T = 1;
  cin >> T;
  for (long long t = 1; 1 < T ? t <= T : t >= T; 1 < T ? t++ : t--) {
    solve();
  }
  return 0;
}

### 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;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class T>
ostream &operator<<(ostream &os, const unordered_set<T> &S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, const pair<L, R> &P) {
  return os << "(" << P.first << "," << P.second << ") ";
}
template <class L, class R>
ostream &operator<<(ostream &os, const map<L, R> &M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
template <class T>
ostream &operator<<(ostream &os, stack<T> S) {
  while (!S.empty()) {
    os << S.top() << "\n";
    S.pop();
  }
  return os;
}
template <class T>
ostream &operator<<(ostream &os, queue<T> Q) {
  while (!Q.empty()) {
    os << Q.front() << " ";
    Q.pop();
  }
  return os;
}
const long long mod = 1e9 + 7;
inline long long modulo(long long x) {
  x = x % mod;
  return x < 0 ? x + mod : x;
}
inline long long add(long long x, long long y) {
  x += y;
  return modulo(x);
}
inline long long sub(long long x, long long y) {
  x -= y;
  return modulo(x);
}
inline long long mul(long long x, long long y) {
  x = x * 1ll * y;
  return modulo(x);
}
inline long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long temp = power(x, y / 2);
  temp = mul(temp, temp);
  if (y % 2 == 1) temp = mul(temp, x);
  return temp;
}
inline long long inv(long long x) { return power(x, mod - 2); }
long long makeall(vector<char> &cache, long long i, long long j, char c) {
  long long count = 0;
  for (long long k = i; i < j ? k <= j : k >= j; i < j ? k++ : k--) {
    if (cache[k] != c) count++;
  }
  return count;
}
long long make(vector<char> &cache, long long i, long long j, char c) {
  if (i == j) return (cache[i] == c) ? 0 : 1;
  long long mid = (i + j) / 2;
  long long temp = makeall(cache, i, mid, c) + make(cache, mid + 1, j, c + 1);
  temp = min(makeall(cache, mid + 1, j, c) + make(cache, i, mid, c + 1), temp);
  return temp;
}
void solve() {
  long long n;
  cin >> n;
  vector<char> cache(n, '@');
  for (long long i = 0; i < n; i++) cin >> cache[i];
  cout << make(cache, 0, n - 1, 'a') << endl;
}
int32_t main() {
  long long T = 1;
  cin >> T;
  for (long long t = 1; 1 < T ? t <= T : t >= T; 1 < T ? t++ : t--) {
    solve();
  }
  return 0;
}

```