#include <bits/stdc++.h>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
using vint = vector<int>;
using vpii = vector<pii>;
pii operator+(pii a, pii b) { return {a.first + b.first, a.second + b.second}; }
pll operator+(pll a, pll b) { return {a.first + b.first, a.second + b.second}; }
namespace IO {
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 24], *O = obuf;
inline ll read() {
  ll x = 0;
  bool sign = 0;
  char s =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (!isdigit(s))
    sign |= s == '-',
        s = (p1 == p2 &&
                     (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
                 ? EOF
                 : *p1++);
  while (isdigit(s))
    x = (x << 1) + (x << 3) + (s - '0'),
    s = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
             ? EOF
             : *p1++);
  return sign ? -x : x;
}
inline void print(ll x) {
  if (x < 0)
    (*O++ = '-'), print(-x);
  else {
    if (x > 9) print(x / 10);
    (*O++ = x % 10 + '0');
  }
}
}  // namespace IO
const int mod = 998244353;
const long double Pi = acos(-1);
namespace math {
const int MAXN = 2e6 + 5;
ll ksm(ll a, ll b, ll p) {
  ll s = 1, m = a;
  while (b) {
    if (b & 1) s = s * m % p;
    m = m * m % p, b >>= 1;
  }
  return s;
}
ll ksm(ll a, ll b) {
  ll s = 1, m = a;
  while (b) {
    if (b & 1) s = s * m % mod;
    m = m * m % mod, b >>= 1;
  }
  return s;
}
ll inv(ll x) { return ksm(x, mod - 2); }
ll fc[MAXN], ifc[MAXN];
void init(int n) {
  assert(n < MAXN);
  fc[0] = 1;
  for (int i = 1; i <= n; i++) fc[i] = fc[i - 1] * i % mod;
  ifc[n] = inv(fc[n]);
  for (int i = n - 1; i >= 0; i--) ifc[i] = ifc[i + 1] * (i + 1) % mod;
}
ll bin(ll n, ll m) { return fc[n] * ifc[m] % mod * ifc[n - m] % mod; }
}  // namespace math
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
using namespace IO;
const int N = 1e3 + 5;
void solve() {
  int n;
  string s, t;
  cin >> n >> s, t = s;
  sort(t.begin(), t.end());
  int cnt = 0;
  for (int i = 0; i < n; i++) cnt += s[i] != t[i];
  cout << cnt << endl;
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
using vint = vector<int>;
using vpii = vector<pii>;
pii operator+(pii a, pii b) { return {a.first + b.first, a.second + b.second}; }
pll operator+(pll a, pll b) { return {a.first + b.first, a.second + b.second}; }
namespace IO {
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 24], *O = obuf;
inline ll read() {
  ll x = 0;
  bool sign = 0;
  char s =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (!isdigit(s))
    sign |= s == '-',
        s = (p1 == p2 &&
                     (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
                 ? EOF
                 : *p1++);
  while (isdigit(s))
    x = (x << 1) + (x << 3) + (s - '0'),
    s = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
             ? EOF
             : *p1++);
  return sign ? -x : x;
}
inline void print(ll x) {
  if (x < 0)
    (*O++ = '-'), print(-x);
  else {
    if (x > 9) print(x / 10);
    (*O++ = x % 10 + '0');
  }
}
}  // namespace IO
const int mod = 998244353;
const long double Pi = acos(-1);
namespace math {
const int MAXN = 2e6 + 5;
ll ksm(ll a, ll b, ll p) {
  ll s = 1, m = a;
  while (b) {
    if (b & 1) s = s * m % p;
    m = m * m % p, b >>= 1;
  }
  return s;
}
ll ksm(ll a, ll b) {
  ll s = 1, m = a;
  while (b) {
    if (b & 1) s = s * m % mod;
    m = m * m % mod, b >>= 1;
  }
  return s;
}
ll inv(ll x) { return ksm(x, mod - 2); }
ll fc[MAXN], ifc[MAXN];
void init(int n) {
  assert(n < MAXN);
  fc[0] = 1;
  for (int i = 1; i <= n; i++) fc[i] = fc[i - 1] * i % mod;
  ifc[n] = inv(fc[n]);
  for (int i = n - 1; i >= 0; i--) ifc[i] = ifc[i + 1] * (i + 1) % mod;
}
ll bin(ll n, ll m) { return fc[n] * ifc[m] % mod * ifc[n - m] % mod; }
}  // namespace math
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
using namespace IO;
const int N = 1e3 + 5;
void solve() {
  int n;
  string s, t;
  cin >> n >> s, t = s;
  sort(t.begin(), t.end());
  int cnt = 0;
  for (int i = 0; i < n; i++) cnt += s[i] != t[i];
  cout << cnt << endl;
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```