4
10 4
15 1
19 3
20 1
 2
1 999999999
1000000000 1000000000
 67
1 1
3 8
4 10
7 8
9 2
10 1
11 5
12 8
13 4
16 6
18 3
19 3
22 5
24 6
27 5
28 3
29 3
30 5
32 5
33 10
34 7
35 8
36 5
41 3
42 2
43 5
46 4
48 4
49 9
52 4
53 9
55 1
56 4
59 7
68 7
69 4
71 9
72 10
74 5
76 4
77 9
80 7
81 9
82 5
83 5
84 9
85 7
86 9
87 4
88 7
89 10
90 3
91 5
92 10
93 5
94 8
95 4
96 2
97 10
98 1
99 3
100 1
101 5
102 4
103 8
104 8
105 8
 10
999999900 1000000000
999999901 1000000000
999999902 1000000000
999999903 1000000000
999999904 1000000000
999999905 1000000000
999999906 1000000000
999999907 1000000000
999999908 1000000000
999999909 1000000000
 35
1 7
3 11
6 12
7 6
8 5
9 11
15 3
16 10
22 2
23 3
25 7
27 3
34 5
35 10
37 3
39 4
40 5
41 1
44 1
47 7
48 11
50 6
52 5
57 2
58 7
60 4
62 1
67 3
68 12
69 8
70 1
71 5
72 5
73 6
74 4
 1
1000000000 1000000000
 2
100000000 1000000000
1000000000 1000000000
 10
7 12
10 2
12 2
15 1
19 2
20 1
53 25
63 10
75 12
87 1
 3
1 1
1000 1000
1000000000 1000000000
 40
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
11 1
12 1
13 1
14 1
15 1
16 1
17 1
18 1
19 1
20 1
21 1
22 1
23 1
24 1
25 1
26 1
27 1
28 1
29 1
30 1
31 1
32 1
33 1
34 1
35 1
36 1
37 1
38 1
39 1
40 1

4
 2
 5
 2
 10
 1
 2
 9
 3
 2

#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long mx = 2e7;
vector<bool> pr(mx, 1);
void sv() {
  pr[1] = 0;
  for (long long n = 2; n * n <= mx; n++) {
    if (pr[n]) {
      for (long long k = 2 * n; k < mx; k += n) pr[k] = 0;
    }
  }
}
long long bit[mx];
void up(long long ind, long long vl) {
  while (ind < mx) {
    bit[ind] += vl;
    ind += ind & (-ind);
  }
}
long long qr(long long ind) {
  long long res = 0;
  while (ind) {
    res += bit[ind];
    ind -= ind & (-ind);
  }
  return res;
}
void solve() {
  long long x;
  scanf("%lld", &(x));
  vector<pair<long long, long long> > s(x);
  for (int n = 0; n < x; n++)
    scanf("%lld", &(s[n].first)), scanf("%lld", &(s[n].second));
  if (x == 1) {
    printf("%lld\n", (x));
    return;
  }
  long long res = 2;
  long long prv = s[0].first;
  for (long long n = 1; n < x - 1; n++) {
    long long a = s[n].first - s[n].second;
    long long b = s[n].first + s[n].second;
    if (a > prv) {
      res++;
      prv = s[n].first;
      continue;
    } else if (b < s[n + 1].first) {
      res++;
      prv = b;
      continue;
    }
    prv = s[n].first;
  }
  printf("%lld\n", (res));
}
int main() { solve(); }
