Borya's Diagnosis（模拟）

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本文介绍了一个关于模拟算法的问题——Borya如何在遵循特定医生就诊顺序的前提下，找到最快完成所有诊断的方法。通过模拟每个医生的工作日程，确定Borya能够完成全部诊疗所需的最短时间。

Borya's Diagnosis

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 s_i-th day and works every d_i day. So, he works on days s_i, s_i + d_i, s_i + 2d_i, ....

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 s_i and d_i (1 ≤ s_i, d_i ≤ 1000).

Output

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

Examples

Input

Output

Input

2
10 1
6 5

Output

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.

思路：模拟，因为题目要求只能按照所给的医生顺序进行就医，所以就按照开始工作的天数和间隔天数进行模拟

如果当前医生开始工作的日期大于需要天数d ，d直接等于医生的工作日期，相当于等到了那一天，如果

医生的开始工作日期小于当前天数说明我们错过了，只能再等他上班，每次隔y天，每次加y知道大于了我们消耗的天数d

直接就让d等于它

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    int n,x,y,d = 0;
    scanf("%d",&n);
    for(int i = 0; i < n; i++){
       scanf("%d%d",&x,&y);
       if(d < x) d = x;
       else{
         while(x <= d) x += y;
         d = x;
       }
    }
    printf("%d\n",d);
    return 0;
}