#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long p = 1, data = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    data = data * 10 + (ch ^ 48);
    ch = getchar();
  }
  return p * data;
}
inline long long qpow(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = a * r;
    a = a * a;
    b >>= 1;
  }
  return r;
}
inline long long gcd(long long a, long long b) {
  if (!a || !b) return (!a) ? b : a;
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b)
    x = 1, y = 0;
  else {
    exgcd(b, a % b, y, x);
    y -= x * (a / b);
  }
}
const int mod = 1e9 + 7;
int a[105], dp[105][105][105][2];
int main() {
  int n = read(), o = 0, e = 0;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (a[i] == 0) continue;
    o += a[i] & 1;
    e += (a[i] % 2 == 0);
  }
  memset(dp, 0x3f3f3f3f, sizeof(dp));
  dp[0][0][0][1] = dp[0][0][0][0] = 0;
  o = n / 2 + (n & 1) - o, e = n / 2 - e;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= o; j++)
      for (int k = 0; k <= e; k++) {
        if (a[i] == 0) {
          if (j >= 1)
            dp[i][j][k][0] =
                min(dp[i - 1][j - 1][k][0], dp[i - 1][j - 1][k][1] + 1);
          else
            dp[i][j][k][0] = 0x3f3f3f3f;
          if (k >= 1)
            dp[i][j][k][1] =
                min(dp[i - 1][j][k - 1][1], dp[i - 1][j][k - 1][0] + 1);
          else
            dp[i][j][k][1] = 0x3f3f3f3f;
        } else {
          dp[i][j][k][0] = (a[i] & 1)
                               ? min(dp[i - 1][j][k][0], dp[i - 1][j][k][1] + 1)
                               : 0x3f3f3f3f;
          dp[i][j][k][1] = (a[i] % 2 == 0)
                               ? min(dp[i - 1][j][k][1], dp[i - 1][j][k][0] + 1)
                               : 0x3f3f3f3f;
        }
      }
  int ans = min(dp[n][o][e][0], dp[n][o][e][1]);
  printf("%d\n", ans);
  return 0;
}
