#include <bits/stdc++.h>
using namespace std;
const long long sz = 131077;
const long long mod = 1e9 + 7;
const double PI = acos(-1);
const double eps = 1e-10;
const long long INF = 1e18 + 5;
long long power(long long n, long long k, long long m) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = (ans * n) % m;
    k /= 2;
    n = (n * n) % m;
  }
  return ans;
}
long long n;
string s;
vector<vector<long long> > dp(sz, vector<long long>(26));
long long md(long long i, long long len, char g) {
  if (len == 1) {
    if (s[i] == g)
      return 0;
    else
      return 1;
  }
  long long c1 = 0, c2 = 0;
  for (long long p = i; p < i + len / 2; p++) {
    if (s[p] != g) {
      c1++;
    }
  }
  for (long long p = i + len / 2; p < i + len; p++) {
    if (s[p] != g) {
      c2++;
    }
  }
  long long op1 =
      c1 + md(i + len / 2, len / 2, (char)((((int)g + 1 - 97) % 26)) + 97);
  long long op2 = c2 + md(i, len / 2, (char)((((int)g + 1 - 97) % 26)) + 97);
  return min(op1, op2);
}
void solver() {
  cin >> n;
  cin >> s;
  for (int i = 1; i < int(n + 1); i++) {
    for (int j = 0; j < int(26); j++) {
      dp[i][j] = dp[i - 1][j];
      if (s[i - 1] != char(j + 97)) dp[i][j]++;
    }
  }
  cout << md(0, n, 'a') << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  long long T = 1;
  cin >> T;
  for (int i = 1; i < int(T + 1); i++) {
    clock_t z = clock();
    solver();
    fprintf(stderr, "Total Time: %.3f\n",
            (double)(clock() - z) / CLOCKS_PER_SEC),
        fflush(stderr);
  }
  return 0;
}

### Prompt

Please formulate 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;
const long long sz = 131077;
const long long mod = 1e9 + 7;
const double PI = acos(-1);
const double eps = 1e-10;
const long long INF = 1e18 + 5;
long long power(long long n, long long k, long long m) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = (ans * n) % m;
    k /= 2;
    n = (n * n) % m;
  }
  return ans;
}
long long n;
string s;
vector<vector<long long> > dp(sz, vector<long long>(26));
long long md(long long i, long long len, char g) {
  if (len == 1) {
    if (s[i] == g)
      return 0;
    else
      return 1;
  }
  long long c1 = 0, c2 = 0;
  for (long long p = i; p < i + len / 2; p++) {
    if (s[p] != g) {
      c1++;
    }
  }
  for (long long p = i + len / 2; p < i + len; p++) {
    if (s[p] != g) {
      c2++;
    }
  }
  long long op1 =
      c1 + md(i + len / 2, len / 2, (char)((((int)g + 1 - 97) % 26)) + 97);
  long long op2 = c2 + md(i, len / 2, (char)((((int)g + 1 - 97) % 26)) + 97);
  return min(op1, op2);
}
void solver() {
  cin >> n;
  cin >> s;
  for (int i = 1; i < int(n + 1); i++) {
    for (int j = 0; j < int(26); j++) {
      dp[i][j] = dp[i - 1][j];
      if (s[i - 1] != char(j + 97)) dp[i][j]++;
    }
  }
  cout << md(0, n, 'a') << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  long long T = 1;
  cin >> T;
  for (int i = 1; i < int(T + 1); i++) {
    clock_t z = clock();
    solver();
    fprintf(stderr, "Total Time: %.3f\n",
            (double)(clock() - z) / CLOCKS_PER_SEC),
        fflush(stderr);
  }
  return 0;
}

```