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 Python (3.4.3) (judge verdict: Runtime Error)

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
using namespace std;
int main() {
  int n, m;
  long long ans = 0;
  cin >> n >> m;
  vector<int> a(n), b(n);
  vector<pair<int, int> > ab(n);
  for(int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
  }
  for(int i = 0; i < n; i++) {
    ab[i] = make_pair(a[i], b[i]);
  }
  sort(ab.begin(), ab.end());  
  for(int i = 0; i < n; i++) {
    int buy = min(m, ab[i].second);
    ans += buy * ab[i].first;
    m -= buy;
  }
  
  cout << ans << endl;
}