3
-1 1 -1
 4
45 -1 41 -1
 1
-1
 5
40 -1 44 46 48
 6
43 40 37 34 -1 -1
 7
-1 2 4 -1 4 1 5
 19
23 26 -1 -1 35 38 41 -1 -1 -1 53 -1 59 62 6 7 8 9 -1
 6
-1 2 6 -1 -1 6
 16
3 8 13 18 23 -1 -1 -1 43 48 53 45 -1 -1 -1 -1
 13
25 24 23 22 24 27 -1 33 -1 2 2 2 -1
 12
-1 17 -1 54 -1 64 -1 74 79 84 -1 94
 8
-1 12 14 16 18 20 -1 -1
 13
2 -1 3 1 3 1 -1 1 3 -1 -1 1 1
 5
-1 40 42 -1 46
 2
1000000000 -1
 14
-1 5 3 -1 -1 31 31 31 -1 31 -1 -1 4 7
 3
-1 1000000000 999999999
 7
32 33 34 -1 -1 37 38
 1
1000000000
 10
29 27 -1 23 42 -1 -1 45 -1 -1
 3
39 42 -1
 15
-1 28 -1 32 34 26 -1 26 -1 -1 26 26 26 -1 -1
 3
1000000000 999999999 1000000000
 17
-1 -1 -1 -1 64 68 72 -1 45 46 47 48 49 50 51 52 53
 3
-1 1 1000000000
 2
-1 1000000000
 3
999999999 -1 1000000000
 1
1
 2
1000000000 1000000000
 3
999999999 1000000000 -1
 7
11 8 5 -1 -1 -1 -1
 3
-1 1 2
 2
-1 21
 8
-1 -1 1 7 -1 9 5 2
 11
9 21 17 13 -1 -1 -1 -1 -1 -1 -1
 3
-1 1000000000 -1
 6
-1 6 1 -1 -1 -1
 1
65
 20
-1 32 37 -1 -1 -1 57 -1 -1 40 31 33 -1 -1 39 47 43 -1 35 32
 5
-1 1 7 -1 5
 18
21 19 -1 -1 -1 48 50 -1 54 -1 5 1 -1 -1 -1 37 36 35
 9
42 39 36 33 -1 -1 -1 34 39
 2
-1 -1

1
 1
 1
 1
 1
 3
 2
 2
 2
 3
 2
 1
 6
 1
 1
 3
 1
 1
 1
 2
 1
 2
 2
 2
 2
 1
 2
 1
 1
 1
 2
 2
 1
 3
 3
 1
 2
 1
 5
 2
 4
 2
 1

#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long inf = 2147383647;
const double pi = 2 * acos(0.0);
const double eps = 1e-9;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
inline long long gcd(long long a, long long b) {
  a = ((a) < 0 ? -(a) : (a));
  b = ((b) < 0 ? -(b) : (b));
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
long long ext_gcd(long long A, long long B, long long* X, long long* Y) {
  long long x2, y2, x1, y1, x, y, r2, r1, q, r;
  x2 = 1;
  y2 = 0;
  x1 = 0;
  y1 = 1;
  for (r2 = A, r1 = B; r1 != 0;
       r2 = r1, r1 = r, x2 = x1, y2 = y1, x1 = x, y1 = y) {
    q = r2 / r1;
    r = r2 % r1;
    x = x2 - (q * x1);
    y = y2 - (q * y1);
  }
  *X = x2;
  *Y = y2;
  return r2;
}
inline long long modInv(long long a, long long m) {
  long long x, y;
  ext_gcd(a, m, &x, &y);
  if (x < 0) x += m;
  return x;
}
inline long long power(long long a, long long p) {
  long long res = 1, x = a;
  while (p) {
    if (p & 1) res = (res * x);
    x = (x * x);
    p >>= 1;
  }
  return res;
}
inline long long bigmod(long long a, long long p, long long m) {
  long long res = 1 % m, x = a % m;
  while (p) {
    if (p & 1) res = (res * x) % m;
    x = (x * x) % m;
    p >>= 1;
  }
  return res;
}
using namespace std;
template <typename T>
ostream& operator<<(ostream& out, vector<T> v) {
  out << "[ ";
  for (int i = 0; i < (((int)(v).size())); i++) {
    if (i) out << ", ";
    out << v[i];
  }
  out << " ]";
  return out;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  out << "( " << p.first << ", " << p.second << ")";
  return out;
}
template <typename T>
ostream& operator,(ostream& out, T x) {
  out << x << " ";
  return out;
}
int N;
long long ara[200007];
int dp[200007];
int rec(int st, int ed) {
  if (st == ed) {
    return 1;
  }
  if (st > ed) {
    return 0;
  }
  int& ret = dp[st];
  if (ret != -1) {
    return ret;
  }
  int a = -1, b = -1;
  for (int i = st; i <= ed; i++) {
    if (ara[i] != -1) {
      if (a == -1) {
        a = i;
      } else {
        b = i;
        break;
      }
    }
  }
  if (a == -1 || b == -1) return ret = 1;
  ret = (1 << 29);
  if ((ara[a] - ara[b]) % (a - b) != 0) {
    return ret = 1 + rec(b, ed);
  }
  long long d = (ara[a] - ara[b]) / (a - b);
  int ret2 = 1;
  long long cur = ara[a];
  for (int i = a - 1; i >= st; i--) {
    cur -= d;
    if (cur <= 0) {
      ret2++;
      ret = 1 + rec(b, ed);
      break;
    }
  }
  cur = ara[b];
  for (int i = b + 1; i <= ed; i++) {
    cur += d;
    if (cur <= 0 || (ara[i] != -1 && ara[i] != cur)) {
      ret2 += rec(i, ed);
      break;
    }
  }
  ret = ((ret) < (ret2) ? (ret) : (ret2));
  return ret;
}
int solve() {
  memset(dp, -1, sizeof(dp));
  return rec(0, N - 1);
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < (N); i++) {
    scanf("%lld", &ara[i]);
  }
  printf("%d\n", solve());
  return 0;
}
