code forces 443 A.Borya's Diagnosis（找规律）

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

32 21 22 2

output

4

input

210 16 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.

【题解】

 这题就是把每个医生的最小上班不重合时间求出来，然后输出其中的最大值即可。

【AC代码】

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