Borya’s Diagnosis

Borya’s Diagnosis

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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.

忽略了一点！！！当sum小于a[i]的时候，不能看病的。。。wa到这里了

ac代码：

#include<iostream>
using namespace std;
long long int n,a[1005],b[1005],c[1005];
int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i]>>b[i];
		c[i]=a[i]%b[i];
	}
	long long int sum=a[0]+1;
	if(n!=1){
		for(int i=1;i<n;i++){
		while(c[i]!=sum%b[i]) sum++;
		if(sum<=a[i]) sum=a[i];//！！！！！！
		//cout<<sum<<endl;
		sum++; 
		
		
	}
	sum--;
	}
	else sum--;
	cout<<sum<<endl;
	return 0;
} 
/*
12
4 5
6 4
8 4
2 4
9 4
5 3
5 7
8 3
8 4
11 34
65 2
3 1
*/