C

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.

There are n trees located along the road at points with coordinates x₁, x₂, ..., x_n. Each tree has its height h_i. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [x_i - h_i, x_i] or [x_i;x_i + h_i]. The tree that is not cut down occupies a single point with coordinate x_i. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of trees.

Next n lines contain pairs of integers x_i, h_i (1 ≤ x_i, h_i ≤ 10⁹) — the coordinate and the height of the і-th tree.

The pairs are given in the order of ascending x_i. No two trees are located at the point with the same coordinate.

Output

Print a single number — the maximum number of trees that you can cut down by the given rules.

Sample Input

Input

5
1 2
2 1
5 10
10 9
19 1

Output

3

Input

5
1 2
2 1
5 10
10 9
20 1

Output

4

Hint

In the first sample you can fell the trees like that:

• fell the 1-st tree to the left — now it occupies segment [ - 1;1]
• fell the 2-nd tree to the right — now it occupies segment [2;3]
• leave the 3-rd tree — it occupies point 5
• leave the 4-th tree — it occupies point 10
• fell the 5-th tree to the right — now it occupies segment [19;20]

In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].

题意：一条路上有树，树有高度h，我们可以将树向左或向右砍倒只要它倒下去不会压到其他树（不论这些树是站着还是倒下了）。求最多可以砍倒多少树。

这一场cf真是水的一匹，我当时怎么没打！！！简直上分福利局。

我们只要按照题目给定的操作走一遍就可以了。优先向左倒，就连数据都已经给你排好序了。

附AC代码：

#include<iostream>
using namespace std;

long long x[100010];
long long h[100010];

int main(){
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>x[i]>>h[i];
    }
    if(n>2){
        int ans=2;
    for(int i=1;i<n-1;i++){
        if(x[i]-h[i]>x[i-1]){
            ans++;
        }
        else if(x[i]+h[i]<x[i+1]){
            ans++;
            x[i]+=h[i];
        }
    }
    cout<<ans<<endl;
    }
    else
    cout<<n<<endl;
    return 0;
}