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.
#Winners never quit, quiters never win............................................................................
from collections import deque as de
import math
from collections import Counter as cnt
from functools import reduce
from typing import MutableMapping

def factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0

arrStack = My_stack()    


def isPrime(n) : 
	if (n <= 1) : 
		return False
	if (n <= 3) : 
		return True

	
	if (n % 2 == 0 or n % 3 == 0) : 
		return False

	i = 5
	while(i * i <= n) : 
		if (n % i == 0 or n % (i + 2) == 0) : 
			return False
		i = i + 6

	return True

def get_prime_factors(number):
    prime_factors = []
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i

    if number > 2:
        prime_factors.append(int(number))

    return prime_factors
def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic
def Log2(x): 
    return (math.log10(x) / 
            math.log10(2)); 

def isPowerOfTwo(n): 
    return (math.ceil(Log2(n)) == math.floor(Log2(n))); 

   
#here we go......................
#winners never quit, quitters never win
n=int(input())
tempp=n-1
lastday=0
while n :
    n-=1
    s,d=map(int,input().split())

    if n==tempp:
        lastday=s
    else:
        #print("executed")
        if s > lastday:
            lastday=s
        elif s == lastday:
            lastday=s+d
        else:
            nn=(lastday+1-s)
            #print("nn=",nn,"d=",d)
            if nn%d==0:
                #print("balbla")
                temp=nn//d
                lastday=s+(temp*d)
            else:
                #print("dontknow")
                temp=nn//d
                temp+=1
                lastday=s+(temp*d)
    #print(lastday)
print(lastday)
            
        