Problem:
Score : 100 points

Problem Statement

In your garden, there is a long and narrow flowerbed that stretches infinitely to the east. You have decided to plant N kinds of flowers in this empty flowerbed. For convenience, we will call these N kinds of flowers Flower 1, 2, …, N. Also, we will call the position that is p centimeters from the west end of the flowerbed Position p.

You will plant Flower i (1 ≤ i ≤ N) as follows: first, plant one at Position w_i, then plant one every d_i centimeters endlessly toward the east. That is, Flower i will be planted at the positions w_i, w_i + d_i, w_i + 2 d_i, … Note that more than one flower may be planted at the same position.

Find the position at which the K-th flower from the west is planted. If more than one flower is planted at the same position, they are counted individually.

Constraints

1 ≤ N ≤ 10^5

1 ≤ K ≤ 10^9

1 ≤ w_i ≤ 10^{18}

1 ≤ d_i ≤ 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
w_1 d_1
:
w_N d_N

Output

When the K-th flower from the west is planted at Position X, print the value of X. (The westmost flower is counted as the 1-st flower.)

Sample Input 1

2 6
20 10
25 15

Sample Output 1

50

Two kinds of flowers are planted at the following positions:

Flower 1: Position 20, 30, 40, 50, 60, …

Flower 2: Position 25, 40, 55, 70, 85, …

The sixth flower from the west is the Flower 1 planted at Position 50. Note that the two flowers planted at Position 40 are counted individually.

Sample Input 2

3 9
10 10
10 10
10 10

Sample Output 2

30

Three flowers are planted at each of the positions 10, 20, 30, … Thus, the ninth flower from the west is planted at Position 30.

Sample Input 3

1 1000000000
1000000000000000000 1000000000

Sample Output 3

1999999999000000000

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define fbo find_by_order
#define ook order_of_key
 
typedef long long ll;
typedef pair<ll,ll> ii;
typedef vector<int> vi;
typedef long double ld; 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
typedef set<int>::iterator sit;
typedef map<int,int>::iterator mit;
typedef vector<int>::iterator vit;

ll w[111111];
ll d[111111];

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	ll n,k; cin>>n>>k;
	for(int i=0;i<n;i++)
	{
		cin>>w[i]>>d[i];
	}
	ll lo = 1; ll hi = ll(4e18);
	ll ans=0;
	while(lo<=hi)
	{
		ll mid=(lo+hi)>>1;
		ll cnt=0;
		for(int i=0;i<n;i++)
		{
			if(w[i]<=mid)
			{
				cnt+=1+(mid-w[i])/d[i];
			}
		}
		if(cnt>=k)
		{
			ans=mid;
			hi=mid-1;
		}
		else
		{
			lo=mid+1;
		}
	}
	cout<<ans<<'\n';
}