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

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class PopulationSize {
    void solve() {
        int n = in.nextInt();
        long[] a = new long[n];
        for (int i = 0; i < n; i++) a[i] = in.nextInt();
        
        int ans = 0;
        for (int i = 0; i < n; ) {
            ans++;
            
            // look for two fixed number a[p] and a[q] where p<q
            int p = i;
            while (p < n && a[p] == -1) p++;
            if (p == n) break;
            
            int q = p + 1;
            while (q < n && a[q] == -1) q++;
            if (q == n) break;
            
            // when we get two fixed numbers, the arithmetic progression is determined. check the part between (p,q) is valid or not
            int steps = q - p;
            if ((a[q] - a[p]) % steps != 0) {
                i = q;
                continue;
            }
            
            // check if the part before 'p' is valid or not
            long d = (a[q] - a[p]) / steps;
            if (d > 0 && a[p] - d * (p - i) <= 0) {
                i = q;
                continue;
            }
            
            // extend the current arithmetic progression as long as possible (since we will get a no worse result by extending it)
            int j = q + 1;
            while (j < n) {
                long expected = a[q] + (j - q) * d;
                if (expected <= 0) break;
                if (a[j] == -1 || a[j] == expected) j++;
                else break;
            }
            
            i = j;
        }
        
        out.println(ans);
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new PopulationSize().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
