#include <bits/stdc++.h>
using namespace std;
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second <= b.second) return true;
  return false;
}
bool prime[10000001];
void sieve() {
  memset(prime, true, sizeof(prime));
  prime[1] = false;
  for (long long i = 2; i * i <= 10000001; i++) {
    if (prime[i] == true) {
      for (long long j = i * i; j <= 10000001; j += i) prime[j] = false;
    }
  }
}
int sprimefactor[1000001];
void smallestprimef() {
  for (long long i = 1; i <= 1000000; i++) sprimefactor[i] = i;
  for (long long i = 2; i * i <= 1000000; i++) {
    if (sprimefactor[i] == i) {
      for (long long j = i * i; j <= 1000000; j++) {
        if (sprimefactor[j] == j) sprimefactor[j] = i;
      }
    }
  }
}
long long egcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long egc = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return egc;
}
long long phi[1000001];
void ph() {
  for (long long i = 1; i < 1000001; i++) phi[i] = i;
  for (long long i = 2; i < 1000001; i++) {
    if (phi[i] == i) {
      phi[i] = i - 1;
      for (long long j = 2 * i; j < 1000001; j += i) {
        phi[j] /= i;
        phi[j] *= (i - 1);
      }
    }
  }
}
long long sumofdigits(long long n) {
  long long sum = 0;
  while (n > 0) {
    sum += (n % 10);
    n /= 10;
  }
  return sum;
}
string s;
long long solve(long long low, long long high, char ch) {
  if (low == high) {
    if (s[low] == ch) return 0;
    return 1;
  }
  long long c1 = 0, c2 = 0, i;
  long long mid = (low + high) / 2;
  for (i = low; i < mid + 1; i++)
    if (s[i] != ch) c1++;
  for (i = mid + 1; i < high + 1; i++)
    if (s[i] != ch) c2++;
  return min(c1 + solve(mid + 1, high, ch + 1), c2 + solve(low, mid, ch + 1));
}
long long i, j, n, x, y;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    long long c = 0, o = 0, e = 0, flag = 0, pos = -1, x = 0, y = 0, z = 0,
              i = 0, j = 0, k = 0;
    long long mini = INT_MAX, maxi = INT_MIN;
    long long n;
    cin >> n;
    cin >> s;
    long long ans = solve(0, n - 1, 'a');
    cout << ans << "\n";
  }
  return 0;
}

### 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;
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second <= b.second) return true;
  return false;
}
bool prime[10000001];
void sieve() {
  memset(prime, true, sizeof(prime));
  prime[1] = false;
  for (long long i = 2; i * i <= 10000001; i++) {
    if (prime[i] == true) {
      for (long long j = i * i; j <= 10000001; j += i) prime[j] = false;
    }
  }
}
int sprimefactor[1000001];
void smallestprimef() {
  for (long long i = 1; i <= 1000000; i++) sprimefactor[i] = i;
  for (long long i = 2; i * i <= 1000000; i++) {
    if (sprimefactor[i] == i) {
      for (long long j = i * i; j <= 1000000; j++) {
        if (sprimefactor[j] == j) sprimefactor[j] = i;
      }
    }
  }
}
long long egcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long egc = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return egc;
}
long long phi[1000001];
void ph() {
  for (long long i = 1; i < 1000001; i++) phi[i] = i;
  for (long long i = 2; i < 1000001; i++) {
    if (phi[i] == i) {
      phi[i] = i - 1;
      for (long long j = 2 * i; j < 1000001; j += i) {
        phi[j] /= i;
        phi[j] *= (i - 1);
      }
    }
  }
}
long long sumofdigits(long long n) {
  long long sum = 0;
  while (n > 0) {
    sum += (n % 10);
    n /= 10;
  }
  return sum;
}
string s;
long long solve(long long low, long long high, char ch) {
  if (low == high) {
    if (s[low] == ch) return 0;
    return 1;
  }
  long long c1 = 0, c2 = 0, i;
  long long mid = (low + high) / 2;
  for (i = low; i < mid + 1; i++)
    if (s[i] != ch) c1++;
  for (i = mid + 1; i < high + 1; i++)
    if (s[i] != ch) c2++;
  return min(c1 + solve(mid + 1, high, ch + 1), c2 + solve(low, mid, ch + 1));
}
long long i, j, n, x, y;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    long long c = 0, o = 0, e = 0, flag = 0, pos = -1, x = 0, y = 0, z = 0,
              i = 0, j = 0, k = 0;
    long long mini = INT_MAX, maxi = INT_MIN;
    long long n;
    cin >> n;
    cin >> s;
    long long ans = solve(0, n - 1, 'a');
    cout << ans << "\n";
  }
  return 0;
}

```