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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i,s,n) for(int i = s; i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(n) (n).begin(), (n).end()
#define RALL(n) (n).rbegin(), (n).rend()
#define ATYN(n) cout << ( (n) ? "Yes":"No") << '\n';
#define CFYN(n) cout << ( (n) ? "YES":"NO") << '\n';
#define OUT(n) cout << (n) << '\n';
using ll = long long;
using ull = unsigned long long;
using pii = pair<int,int>;
using pll = pair<ll,ll>;

int main(void)
{
    IOS
    ll n, k;
    cin >> n >> k;
    vector<ll> w(n), d(n);
    REP(i,n) cin >> w[i] >> d[i];

    auto isok = [&](ll x) {
        ll f = 0;
        REP(i,n) {
            if (x < w[i]) continue;
            f += (x-w[i]) / d[i]+1;
            if (k <= f) return true;
        }
        //cerr << f << endl;
        return (k <= f);
    };

    ll ok = 1LL << 62, ng = 0;

    while(abs(ok - ng) > 1) {
        ll mid = (ok + ng) / 2;
        if (isok(mid)) ok = mid;
        else ng = mid;
        //cerr << mid << endl;
    }

    OUT(ok)

    return 0;
}