source

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,?...,?xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi?-?hi,?xi] or [xi;xi?+?hi]. The tree that is not cut down occupies a single point with coordinate xi. 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 xi,?hi (1?≤?xi,?hi?≤?109) — the coordinate and the height of the ?-th tree.

The pairs are given in the order of ascending xi. 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.

Example

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

Note

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].

題意：給出n棵樹的位置xi和高度hi，每棵樹可以向左邊砍或則向右邊砍。向左砍則樹的覆蓋位置為xi-hi；向右砍則樹的覆蓋位置為xi+hi。該樹可以砍的條件是不能碰到上（下）一棵樹的覆蓋位置或者上（下）一棵樹的樹的位置。

題解 : 因為對當前樹的不砍 不會對 下一棵樹的操作產生有利的影響，所以砍樹策略是能砍就盡量砍，優先考慮向左砍，再考慮向右砍，不能砍則跳過該樹。

#include <stdio.h>
#include <string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,sum=0;
    pair<int,int> arr[100111];
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&arr[i].first,&arr[i].second);
    }
    arr[0].first=-2147483648;
    arr[n+1].first=2147483647;
    for(int i=1;i<=n;i++)
    {
        if(arr[i].first-arr[i].second>arr[i-1].first)
        {
            sum++;
        }
        else if(arr[i].first+arr[i].second<arr[i+1].first)
        {
            sum++;
            arr[i].first+=arr[i].second;
        }
    }
    printf("%d",sum);

}