#include <bits/stdc++.h>
using namespace std;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
const int mod = (int)1e9 + 7;
const long long INF64 = 3e18;
void smxl(long long &a, long long b) {
  if (a < b) a = b;
}
void smnl(long long &a, long long b) {
  if (a > b) a = b;
}
void adsl(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void misl(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
void smx(long long &a, long long b) {
  if (a < b) a = b;
}
void smn(long long &a, long long b) {
  if (a > b) a = b;
}
void ads(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void mis(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long skr(int a) {
  long long x = (long long)a;
  return a * a;
}
long long kub(int a) {
  long long x = (long long)a;
  return a * a * a;
}
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 d = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long mbinp(long long a, long long b) {
  a %= mod;
  if (b == 0) return 1;
  long long ans = mbinp(a, b / 2);
  long long tmp = (ans * ans) % mod;
  if (b % 2) return ((tmp * a) % mod);
  return ((tmp) % mod);
}
long long binp(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = binp(a, b / 2);
  long long tmp = (ans * ans);
  if (b % 2) return ((tmp * a));
  return ((tmp));
}
long long C(int n, int m) {
  long long ret = 1;
  for (int i = 1; i <= m; i++) {
    ret *= (n - i + 1);
    ret /= i;
  }
  return ret;
}
long long overbinp(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      if (res < INF64 / a)
        res *= a;
      else
        return INF64;
    }
    if (b > 1) {
      if (a < INF64 / a)
        a *= a;
      else
        return INF64;
    }
    b >>= 1;
  }
  return res;
}
int len;
string str;
int solve(int start, int end, char c) {
  int c1, c2;
  c1 = c2 = 0;
  if (end == start + 1) {
    return ((c == str[start] ? 0 : 1));
  }
  int mid = (start + end) / 2;
  for (int i = start; i < mid; i++)
    if (str[i] != c) c1++;
  for (int i = mid; i < end; i++)
    if (str[i] != c) c2++;
  return (min(c1 + solve(mid, end, c + 1), c2 + solve(start, mid, c + 1)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    cin >> len;
    cin >> str;
    cout << solve(0, len, 'a') << '\n';
  }
}

### 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;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
const int mod = (int)1e9 + 7;
const long long INF64 = 3e18;
void smxl(long long &a, long long b) {
  if (a < b) a = b;
}
void smnl(long long &a, long long b) {
  if (a > b) a = b;
}
void adsl(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void misl(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
void smx(long long &a, long long b) {
  if (a < b) a = b;
}
void smn(long long &a, long long b) {
  if (a > b) a = b;
}
void ads(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void mis(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long skr(int a) {
  long long x = (long long)a;
  return a * a;
}
long long kub(int a) {
  long long x = (long long)a;
  return a * a * a;
}
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 d = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long mbinp(long long a, long long b) {
  a %= mod;
  if (b == 0) return 1;
  long long ans = mbinp(a, b / 2);
  long long tmp = (ans * ans) % mod;
  if (b % 2) return ((tmp * a) % mod);
  return ((tmp) % mod);
}
long long binp(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = binp(a, b / 2);
  long long tmp = (ans * ans);
  if (b % 2) return ((tmp * a));
  return ((tmp));
}
long long C(int n, int m) {
  long long ret = 1;
  for (int i = 1; i <= m; i++) {
    ret *= (n - i + 1);
    ret /= i;
  }
  return ret;
}
long long overbinp(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      if (res < INF64 / a)
        res *= a;
      else
        return INF64;
    }
    if (b > 1) {
      if (a < INF64 / a)
        a *= a;
      else
        return INF64;
    }
    b >>= 1;
  }
  return res;
}
int len;
string str;
int solve(int start, int end, char c) {
  int c1, c2;
  c1 = c2 = 0;
  if (end == start + 1) {
    return ((c == str[start] ? 0 : 1));
  }
  int mid = (start + end) / 2;
  for (int i = start; i < mid; i++)
    if (str[i] != c) c1++;
  for (int i = mid; i < end; i++)
    if (str[i] != c) c2++;
  return (min(c1 + solve(mid, end, c + 1), c2 + solve(start, mid, c + 1)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    cin >> len;
    cin >> str;
    cout << solve(0, len, 'a') << '\n';
  }
}

```