Problem:
Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

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

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<n; i++)
#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define all(x) x.begin(),x.end()
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
 
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

//
int main(void) {
    int N,M;
    cin >> N >> M;
    vector<i_i> v(N);
    rep(i,N) cin >> v.at(i).second >> v.at(i).first;
    sort(all(v),
        [](const i_i& x, const i_i& y){return x.second < y.second;});
    ll ans=0;
    int cnt=0;
    int tmp=0;
    while(cnt<M){
        while(cnt<M&&v.at(tmp).first>0){
            ans += v.at(tmp).second;
            v.at(tmp).first--;
            cnt++;
        }
        tmp++;
    }
    cout << ans << endl;
    return 0;
}