Codeforces Round #443 (Div. 2) A. Borya's Diagnosis

http://codeforces.com/problemset/problem/879/A

It seems that Borya is seriously sick. He is going visit n doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor.

Doctors have a strange working schedule. The doctor i goes to work on the si-th day and works every di day. So, he works on days si, si + di, si + 2di, ….

The doctor’s appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors?

Input

First line contains an integer n — number of doctors (1 ≤ n ≤ 1000).

Next n lines contain two numbers si and di (1 ≤ si, di ≤ 1000).

Output

Output a single integer — the minimum day at which Borya can visit the last doctor.

Examples

input

3
2 2
1 2
2 2

output

4

input

2
10 1
6 5

output

11

Note

In the first sample case, Borya can visit all doctors on days 2, 3 and 4.

In the second sample case, Borya can visit all doctors on days 10 and 11.

题意

病人去看医生，医生的上班规律是，第ｓ天上班之后每ｄ天上一次班，问最后一次看医生是第几天。病人看医生也有要求，要求看医生需要按顺序看，也就是，看完第一个医生才能看下一个医生

题解

因为看病有顺序，就方便了很多，看下一个医生的日子是上一个医生之后上班最早的那天，只需一个循环更新一下看完医生的值就行了。

CODE

#include 
#include 
#include 
#include 
#include 
using namespace std;
int n,ans,s,d;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&s,&d);
        for(;s<=ans;s+=d);
            ans=s;
    }
    cout << ans << endl;
}