#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (a < b) swap(a, b);
  return gcd(a % b, b);
}
long long gcdextended(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, gcd = gcdextended(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long modinverse(long long a, long long m) {
  long long x, y;
  long long g = gcdextended(a, m, x, y);
  if (g != 1) {
    cout << "NOT POSSIBLE";
    return -1;
  } else
    return (x % m + m) % m;
}
long long po(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long pom(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long root(long long x, vector<long long> &id) {
  while (id[x] != x) {
    id[x] = id[id[x]];
    x = id[x];
  }
  return x;
}
void uni(long long x, long long y, vector<long long> &id) {
  long long a = root(x, id);
  long long b = root(y, id);
  id[a] = id[b];
}
bool sec(const pair<long long, long long> &a,
         const pair<long long, long long> &b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second < b.second;
}
long long ff(long long l, long long r, string &second, char c) {
  if (l == r) {
    if (second[l] == c) return 0;
    return 1;
  }
  long long i, m = (l + r) / 2, a = 0, b = 0;
  for (i = l; i < m + 1; i++)
    if (second[i] != c) a++;
  for (i = m + 1; i < r + 1; i++)
    if (second[i] != c) b++;
  return min(a + ff(m + 1, r, second, c + 1), b + ff(l, m, second, c + 1));
}
void solve() {
  long long n, i, j;
  cin >> n;
  string second;
  cin >> second;
  cout << ff(0, n - 1, second, 'a') << endl;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (a < b) swap(a, b);
  return gcd(a % b, b);
}
long long gcdextended(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, gcd = gcdextended(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long modinverse(long long a, long long m) {
  long long x, y;
  long long g = gcdextended(a, m, x, y);
  if (g != 1) {
    cout << "NOT POSSIBLE";
    return -1;
  } else
    return (x % m + m) % m;
}
long long po(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long pom(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long root(long long x, vector<long long> &id) {
  while (id[x] != x) {
    id[x] = id[id[x]];
    x = id[x];
  }
  return x;
}
void uni(long long x, long long y, vector<long long> &id) {
  long long a = root(x, id);
  long long b = root(y, id);
  id[a] = id[b];
}
bool sec(const pair<long long, long long> &a,
         const pair<long long, long long> &b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second < b.second;
}
long long ff(long long l, long long r, string &second, char c) {
  if (l == r) {
    if (second[l] == c) return 0;
    return 1;
  }
  long long i, m = (l + r) / 2, a = 0, b = 0;
  for (i = l; i < m + 1; i++)
    if (second[i] != c) a++;
  for (i = m + 1; i < r + 1; i++)
    if (second[i] != c) b++;
  return min(a + ff(m + 1, r, second, c + 1), b + ff(l, m, second, c + 1));
}
void solve() {
  long long n, i, j;
  cin >> n;
  string second;
  cin >> second;
  cout << ff(0, n - 1, second, 'a') << endl;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```