Bear Limak prepares problems for a programming competition. Of course, it would be unprofessional to mention the sponsor name in the statement. Limak takes it seriously and he is going to change some words. To make it still possible to read, he will try to modify each word as little as possible.

Limak has a string s that consists of uppercase English letters. In one move he can swap two adjacent letters of the string. For example, he can transform a string "ABBC" into "BABC" or "ABCB" in one move.

Limak wants to obtain a string without a substring "VK" (i.e. there should be no letter 'V' immediately followed by letter 'K'). It can be easily proved that it's possible for any initial string s.

What is the minimum possible number of moves Limak can do?

Input

The first line of the input contains an integer n (1 ≤ n ≤ 75) — the length of the string.

The second line contains a string s, consisting of uppercase English letters. The length of the string is equal to n.

Output

Print one integer, denoting the minimum possible number of moves Limak can do, in order to obtain a string without a substring "VK".

Examples

Input

4
VKVK


Output

3


Input

5
BVVKV


Output

2


Input

7
VVKEVKK


Output

3


Input

20
VKVKVVVKVOVKVQKKKVVK


Output

8


Input

5
LIMAK


Output

0

Note

In the first sample, the initial string is "VKVK". The minimum possible number of moves is 3. One optimal sequence of moves is:

  1. Swap two last letters. The string becomes "VKKV".
  2. Swap first two letters. The string becomes "KVKV".
  3. Swap the second and the third letter. The string becomes "KKVV". Indeed, this string doesn't have a substring "VK".



In the second sample, there are two optimal sequences of moves. One is "BVVKV" → "VBVKV" → "VVBKV". The other is "BVVKV" → "BVKVV" → "BKVVV".

In the fifth sample, no swaps are necessary.
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0},  {0, -1},
                        {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <typename t, typename t1>
t min1(t a, t1 b) {
  return a < b ? a : b;
}
template <typename t, typename... arg>
t min1(t a, arg... arr) {
  return min1(a, min1(arr...));
}
template <typename t, typename t1>
t max1(t a, t1 b) {
  return a > b ? a : b;
}
template <typename t, typename... arg>
t max1(t a, arg... arr) {
  return max1(a, max1(arr...));
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / max(abs1(a), abs1(b)) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 0) return 0;
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitnum(long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitmaxl(unsigned int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
inline int bitmaxl(long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
  if (mod == -1) mod = 1000000007;
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <class t>
void output1(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) cerr << arr[i] << ' ';
  cerr << endl;
}
template <class t>
void output2(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) output1(arr[i]);
}
const int maxn = 80;
int dp[maxn][maxn][maxn][3];
char str[maxn];
int n;
int arr[maxn], cnt[3];
int bef[3][maxn][3];
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d%s", &n, str);
  for (int i = 0; i < n; i++) {
    if (str[i] == 'V')
      arr[i] = 0;
    else if (str[i] == 'K')
      arr[i] = 1;
    else
      arr[i] = 2;
    for (int j = 0; j < 3; j++) bef[arr[i]][cnt[arr[i]]][j] = cnt[j];
    cnt[arr[i]]++;
  }
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0][0][2] = 0;
  for (int i = 0; i < n; i++) {
    for (int s0 = 0; s0 <= min(i, cnt[0]); s0++)
      for (int s1 = 0; s0 + s1 <= i && s1 <= cnt[1]; s1++)
        for (int last = 0; last < 3; last++) {
          int s2 = i - s0 - s1;
          if (s0 < cnt[0]) {
            int sbef = max(0, bef[0][s0][1] - s1) + max(0, bef[0][s0][2] - s2);
            dp[i + 1][s0 + 1][s1][0] =
                min(dp[i + 1][s0 + 1][s1][0], dp[i][s0][s1][last] + sbef);
          }
          if (s1 < cnt[1] && last != 0) {
            int sbef = max(0, bef[1][s1][0] - s0) + max(0, bef[1][s1][2] - s2);
            dp[i + 1][s0][s1 + 1][1] =
                min(dp[i + 1][s0][s1 + 1][1], dp[i][s0][s1][last] + sbef);
          }
          if (s2 < cnt[2]) {
            int sbef = max(0, bef[2][s2][0] - s0) + max(0, bef[2][s2][1] - s1);
            dp[i + 1][s0][s1][2] =
                min(dp[i + 1][s0][s1][2], dp[i][s0][s1][last] + sbef);
          }
        }
  }
  int ans = 1000000007;
  for (int i = 0; i < 3; i++) ans = min(dp[n][cnt[0]][cnt[1]][i], ans);
  cout << ans << endl;
  return 0;
}
