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>
#define all(v) v.begin(), v.end()
#define mp make_pair
using namespace std;

using ll = long long;
using ii = pair<int, int>;
const int INF = (int)10e9;

int main() {
	std::ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int N, M;
	cin >> N >> M;
	vector<pair<long, long>> v(N);
	for(int i = 0; i < N; i++) {
		ll a, b;
		cin>>a>>b;
		v[i] = {a,b};
	}
	sort(all(v));
	ll res = 0;
	for(int i = 0; i < N; i++) {
		ll can = v[i].second;
		if(M - can >= 0) M -= can;
		else {
			can = M;
			M = 0;
		}
		res += (v[i].first * can);
	}
	cout<<res<<endl;

    return 0;
}