#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T1, class T2>
inline void gn(T1 &x1, T2 &x2) {
  gn(x1), gn(x2);
}
template <class T1, class T2, class T3>
inline void gn(T1 &x1, T2 &x2, T3 &x3) {
  gn(x1), gn(x2), gn(x3);
}
template <class T1, class T2, class T3, class T4>
inline void gn(T1 &x1, T2 &x2, T3 &x3, T4 &x4) {
  gn(x1), gn(x2), gn(x3), gn(x4);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T1, class T2>
inline void print(T1 x1, T2 x2) {
  printsp(x1), println(x2);
}
template <class T1, class T2, class T3>
inline void print(T1 x1, T2 x2, T3 x3) {
  printsp(x1), printsp(x2), println(x3);
}
template <class T1, class T2, class T3, class T4>
inline void print(T1 x1, T2 x2, T3 x3, T4 x4) {
  printsp(x1), printsp(x2), printsp(x3), println(x4);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
char s[100010];
int mx[2][100010], num[100010];
pair<int, int> dp[100010];
int main() {
  int n, m;
  scanf("%d %s %d", &n, s + 1, &m);
  for (int i = 1; i <= n; i++) {
    mx[0][i] = mx[1][i - 1] + 1;
    mx[1][i] = mx[0][i - 1] + 1;
    num[i] = num[i - 1];
    if (s[i] == 'a') mx[1][i] = 0;
    if (s[i] == 'b') mx[0][i] = 0;
    if (s[i] == '?') num[i]++;
  }
  char c = 'b';
  if (m & 1) c = 'a';
  for (int i = m; i <= n; i++) {
    dp[i] = dp[i - 1];
    pair<int, int> p = dp[i - m];
    int na = mx[0][i], nb = mx[1][i];
    if (c == 'a' and na >= m or c == 'b' and nb >= m) {
      p.first++;
      p.second += num[i] - num[i - m];
    }
    if (p.first > dp[i].first)
      dp[i] = p;
    else if (p.first == dp[i].first and p.second < dp[i].second)
      dp[i] = p;
  }
  println(dp[n].second);
}

### Prompt

Please create a solution in cpp to the following problem:

Vasya wrote down two strings s of length n and t of length m consisting of small English letters 'a' and 'b'. What is more, he knows that string t has a form "abab...", namely there are letters 'a' on odd positions and letters 'b' on even positions.

Suddenly in the morning, Vasya found that somebody spoiled his string. Some letters of the string s were replaced by character '?'.

Let's call a sequence of positions i, i + 1, ..., i + m - 1 as occurrence of string t in s, if 1 ≤ i ≤ n - m + 1 and t1 = si, t2 = si + 1, ..., tm = si + m - 1.

The boy defines the beauty of the string s as maximum number of disjoint occurrences of string t in s. Vasya can replace some letters '?' with 'a' or 'b' (letters on different positions can be replaced with different letter). Vasya wants to make some replacements in such a way that beauty of string s is maximum possible. From all such options, he wants to choose one with the minimum number of replacements. Find the number of replacements he should make.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the length of s.

The second line contains the string s of length n. It contains small English letters 'a', 'b' and characters '?' only.

The third line contains a single integer m (1 ≤ m ≤ 105) — the length of t. The string t contains letters 'a' on odd positions and 'b' on even positions.

Output

Print the only integer — the minimum number of replacements Vasya has to perform to make the beauty of string s the maximum possible.

Examples

Input

5
bb?a?
1


Output

2


Input

9
ab??ab???
3


Output

2

Note

In the first sample string t has a form 'a'. The only optimal option is to replace all characters '?' by 'a'.

In the second sample using two replacements we can make string equal to "aba?aba??". It is impossible to get more than two occurrences.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T1, class T2>
inline void gn(T1 &x1, T2 &x2) {
  gn(x1), gn(x2);
}
template <class T1, class T2, class T3>
inline void gn(T1 &x1, T2 &x2, T3 &x3) {
  gn(x1), gn(x2), gn(x3);
}
template <class T1, class T2, class T3, class T4>
inline void gn(T1 &x1, T2 &x2, T3 &x3, T4 &x4) {
  gn(x1), gn(x2), gn(x3), gn(x4);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T1, class T2>
inline void print(T1 x1, T2 x2) {
  printsp(x1), println(x2);
}
template <class T1, class T2, class T3>
inline void print(T1 x1, T2 x2, T3 x3) {
  printsp(x1), printsp(x2), println(x3);
}
template <class T1, class T2, class T3, class T4>
inline void print(T1 x1, T2 x2, T3 x3, T4 x4) {
  printsp(x1), printsp(x2), printsp(x3), println(x4);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
char s[100010];
int mx[2][100010], num[100010];
pair<int, int> dp[100010];
int main() {
  int n, m;
  scanf("%d %s %d", &n, s + 1, &m);
  for (int i = 1; i <= n; i++) {
    mx[0][i] = mx[1][i - 1] + 1;
    mx[1][i] = mx[0][i - 1] + 1;
    num[i] = num[i - 1];
    if (s[i] == 'a') mx[1][i] = 0;
    if (s[i] == 'b') mx[0][i] = 0;
    if (s[i] == '?') num[i]++;
  }
  char c = 'b';
  if (m & 1) c = 'a';
  for (int i = m; i <= n; i++) {
    dp[i] = dp[i - 1];
    pair<int, int> p = dp[i - m];
    int na = mx[0][i], nb = mx[1][i];
    if (c == 'a' and na >= m or c == 'b' and nb >= m) {
      p.first++;
      p.second += num[i] - num[i - m];
    }
    if (p.first > dp[i].first)
      dp[i] = p;
    else if (p.first == dp[i].first and p.second < dp[i].second)
      dp[i] = p;
  }
  println(dp[n].second);
}

```