Codeforces Round #443 (Div. 2) A. Borya's Diagnosis（水水的模拟）

Bertha 。 2022-06-06 02:05 141阅读 0赞

A. Borya's Diagnosis

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 3and 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* + 2*d**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

3
    2 2
    1 2
    2 2

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.

题解：

按题目意思模拟得去就过了

#include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<vector>
    #include<deque>
    #include<algorithm>
    #define ll long long
    #define INF 1008611111
    #define M (t[k].l+t[k].r)/2
    #define lson k*2
    #define rson k*2+1
    using namespace std;
    int main()
    {
        int i,j,n,x,y,d=0;
        scanf("%d",&n);
        for(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",d);
        return 0;
    }