#include <bits/stdc++.h>
const int mod = 1e9 + 7;
using namespace std;
bool sortcol(const vector<int>& v1, const vector<int>& v2) {
  return v1[1] < v2[1];
}
void fastio() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool sortbysecdesc(const pair<long long int, long long int>& a,
                   const pair<long long int, long long int>& b) {
  if (a.first > b.first) return true;
  if (a.first == b.first) return a.second < b.second;
  return false;
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
int fun(int l, int r, int x, int a[]) {
  if (l == r) {
    if (a[l] == x) return 0;
    return 1;
  }
  int t1 = 0, t2 = 0;
  int mid = (l + r) / 2;
  for (int i = l; i <= mid; i++)
    if (a[i] != x) t1++;
  for (int i = mid + 1; i <= r; i++)
    if (a[i] != x) t2++;
  int ans1 = t1 + fun(mid + 1, r, x + 1, a);
  int ans2 = t2 + fun(l, mid, x + 1, a);
  return min(ans1, ans2);
}
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  int a[n + 1];
  for (int i = 0; i < n; i++) a[i + 1] = s[i] - 'a';
  cout << fun(1, n, 0, a) << "\n";
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) solve();
}

### Prompt

Your challenge is to write 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>
const int mod = 1e9 + 7;
using namespace std;
bool sortcol(const vector<int>& v1, const vector<int>& v2) {
  return v1[1] < v2[1];
}
void fastio() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool sortbysecdesc(const pair<long long int, long long int>& a,
                   const pair<long long int, long long int>& b) {
  if (a.first > b.first) return true;
  if (a.first == b.first) return a.second < b.second;
  return false;
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
int fun(int l, int r, int x, int a[]) {
  if (l == r) {
    if (a[l] == x) return 0;
    return 1;
  }
  int t1 = 0, t2 = 0;
  int mid = (l + r) / 2;
  for (int i = l; i <= mid; i++)
    if (a[i] != x) t1++;
  for (int i = mid + 1; i <= r; i++)
    if (a[i] != x) t2++;
  int ans1 = t1 + fun(mid + 1, r, x + 1, a);
  int ans2 = t2 + fun(l, mid, x + 1, a);
  return min(ans1, ans2);
}
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  int a[n + 1];
  for (int i = 0; i < n; i++) a[i + 1] = s[i] - 'a';
  cout << fun(1, n, 0, a) << "\n";
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) solve();
}

```