#include <bits/stdc++.h>
using namespace std;
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const int iINF = INT_MAX;
const long long INF = LLONG_MAX;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fpow(long long a, long long b, long long m) {
  if (!b) return 1;
  long long ans = fpow(a * a % m, b / 2, m);
  return (b % 2 ? ans * a % m : ans);
}
long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
void solve() {
  int n;
  string s;
  cin >> n >> s;
  int alp[26];
  for (long long i = 0; i < 26; i++) alp[i] = 0;
  for (long long i = 0; i < n; i++) {
    int k = s[i] - 'a';
    alp[k]++;
  }
  string a = "";
  for (long long i = 0; i < 26; i++) {
    for (long long j = 0; j < alp[i]; j++) a += (i + 'a');
  }
  int cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (s[i] != a[i]) cnt++;
  }
  cout << cnt << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

A string s of length n, consisting of lowercase letters of the English alphabet, is given.

You must choose some number k between 0 and n. Then, you select k characters of s and permute them however you want. In this process, the positions of the other n-k characters remain unchanged. You have to perform this operation exactly once.

For example, if s="andrea", you can choose the k=4 characters "a_d_ea" and permute them into "d_e_aa" so that after the operation the string becomes "dneraa".

Determine the minimum k so that it is possible to sort s alphabetically (that is, after the operation its characters appear in alphabetical order).

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 40) — the length of the string.

The second line of each test case contains the string s. It is guaranteed that s contains only lowercase letters of the English alphabet.

Output

For each test case, output the minimum k that allows you to obtain a string sorted alphabetically, through the operation described above.

Example

Input


4
3
lol
10
codeforces
5
aaaaa
4
dcba


Output


2
6
0
4

Note

In the first test case, we can choose the k=2 characters "_ol" and rearrange them as "_lo" (so the resulting string is "llo"). It is not possible to sort the string choosing strictly less than 2 characters.

In the second test case, one possible way to sort s is to consider the k=6 characters "_o__force_" and rearrange them as "_c__efoor_" (so the resulting string is "ccdeefoors"). One can show that it is not possible to sort the string choosing strictly less than 6 characters.

In the third test case, string s is already sorted (so we can choose k=0 characters).

In the fourth test case, we can choose all k=4 characters "dcba" and reverse the whole string (so the resulting string is "abcd").

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const int iINF = INT_MAX;
const long long INF = LLONG_MAX;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fpow(long long a, long long b, long long m) {
  if (!b) return 1;
  long long ans = fpow(a * a % m, b / 2, m);
  return (b % 2 ? ans * a % m : ans);
}
long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
void solve() {
  int n;
  string s;
  cin >> n >> s;
  int alp[26];
  for (long long i = 0; i < 26; i++) alp[i] = 0;
  for (long long i = 0; i < n; i++) {
    int k = s[i] - 'a';
    alp[k]++;
  }
  string a = "";
  for (long long i = 0; i < 26; i++) {
    for (long long j = 0; j < alp[i]; j++) a += (i + 'a');
  }
  int cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (s[i] != a[i]) cnt++;
  }
  cout << cnt << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
}

```