#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long MOD = 998244353;
const long long inf = 1e18;
const long long MAX = 2e5 + 1;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long pwr(long long x, long long n) {
  x = x % mod;
  if (!n) return 1;
  if (n & 1)
    return mul(x, pwr(mul(x, x), (n - 1) / 2));
  else
    return pwr(mul(x, x), n / 2);
}
long long modinv(long long n) { return pwr(n, mod - 2); }
long long inv(long long i) {
  if (i == 1) return 1;
  return (MOD - (MOD / i) * inv(MOD % i) % MOD) % MOD;
}
bool f(int a, int b) { return a > b; }
struct cmp {
  bool operator()(const int &k1, const int &k2) { return k1 > k2; }
};
int n;
string s;
long long answer(long long low, long long high, char c) {
  if (low == high) {
    if (s[low] == c)
      return 0;
    else
      return 1;
  }
  long long mid = (low + high) / 2;
  long long left = 0, right = 0;
  for (long long i = low; i <= high; i++) {
    if (s[i] != c) {
      if (i <= mid)
        left++;
      else
        right++;
    }
  }
  return min(left + answer(mid + 1, high, c + 1),
             right + answer(low, mid, c + 1));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  cin >> tt;
  while (tt--) {
    cin >> n >> s;
    long long ans = answer(0, n - 1, 'a');
    cout << ans << "\n";
  }
  return 0;
}

### 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;
const long long mod = 1000000007;
const long long MOD = 998244353;
const long long inf = 1e18;
const long long MAX = 2e5 + 1;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long pwr(long long x, long long n) {
  x = x % mod;
  if (!n) return 1;
  if (n & 1)
    return mul(x, pwr(mul(x, x), (n - 1) / 2));
  else
    return pwr(mul(x, x), n / 2);
}
long long modinv(long long n) { return pwr(n, mod - 2); }
long long inv(long long i) {
  if (i == 1) return 1;
  return (MOD - (MOD / i) * inv(MOD % i) % MOD) % MOD;
}
bool f(int a, int b) { return a > b; }
struct cmp {
  bool operator()(const int &k1, const int &k2) { return k1 > k2; }
};
int n;
string s;
long long answer(long long low, long long high, char c) {
  if (low == high) {
    if (s[low] == c)
      return 0;
    else
      return 1;
  }
  long long mid = (low + high) / 2;
  long long left = 0, right = 0;
  for (long long i = low; i <= high; i++) {
    if (s[i] != c) {
      if (i <= mid)
        left++;
      else
        right++;
    }
  }
  return min(left + answer(mid + 1, high, c + 1),
             right + answer(low, mid, c + 1));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  cin >> tt;
  while (tt--) {
    cin >> n >> s;
    long long ans = answer(0, n - 1, 'a');
    cout << ans << "\n";
  }
  return 0;
}

```