#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int sqr(long long int n) { return n * n; }
long long int mod = 1e9 + 7;
long long int pow(long long int p, long long int d) {
  return d ? (pow((p * p) % mod, d >> 1) * (d & 1 ? p : 1)) % mod : 1ll;
}
vector<bool> SieveOfEratosthenes(long long int num) {
  vector<bool> soe;
  soe.assign(num + 1, true);
  soe[0] = false;
  soe[1] = false;
  for (long long int i = 2; i * i <= num; i++) {
    if (soe[i] == true) {
      for (long long int j = i * 2; j <= num; j += i) soe[j] = false;
    }
  }
  return soe;
}
long long int bs(vector<long long int> &v, long long int l, long long int r,
                 long long int key) {
  while (l + 1 < r) {
    long long int mid = (l + r) / 2;
    if (v[mid] > key)
      r = mid;
    else
      l = mid;
  }
  return l;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string s;
    cin >> s;
    ;
    string k = s;
    sort(k.begin(), k.end());
    long long int cnt = 0;
    for (int i = 0; i < s.size(); ++i) {
      if (s[i] != k[i]) cnt++;
    }
    cout << cnt << '\n';
  }
}

### Prompt

In cpp, your task is to solve 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;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int sqr(long long int n) { return n * n; }
long long int mod = 1e9 + 7;
long long int pow(long long int p, long long int d) {
  return d ? (pow((p * p) % mod, d >> 1) * (d & 1 ? p : 1)) % mod : 1ll;
}
vector<bool> SieveOfEratosthenes(long long int num) {
  vector<bool> soe;
  soe.assign(num + 1, true);
  soe[0] = false;
  soe[1] = false;
  for (long long int i = 2; i * i <= num; i++) {
    if (soe[i] == true) {
      for (long long int j = i * 2; j <= num; j += i) soe[j] = false;
    }
  }
  return soe;
}
long long int bs(vector<long long int> &v, long long int l, long long int r,
                 long long int key) {
  while (l + 1 < r) {
    long long int mid = (l + r) / 2;
    if (v[mid] > key)
      r = mid;
    else
      l = mid;
  }
  return l;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string s;
    cin >> s;
    ;
    string k = s;
    sort(k.begin(), k.end());
    long long int cnt = 0;
    for (int i = 0; i < s.size(); ++i) {
      if (s[i] != k[i]) cnt++;
    }
    cout << cnt << '\n';
  }
}

```