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 rept(i,a,b) for(int i=a; i<b; i++)
#define all(x) (x).begin(),(x).end()

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> ii;

int main() {
	ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

	ll n, m;
	cin >> n >> m;

	vector< pair<ll,ll> > v(n);
	vi a_sort(n);
	rep(i, n){
		cin >> v[i].first >> v[i].second;
	}

	sort(all(v));

	ll ans = 0;
	rep(j, n){
		if(m > v[j].second){
			ans += v[j].first * v[j].second;
			m = m - v[j].second;
		}else{
			ans += v[j].first * m;
			m = 0;
		}
	}


	cout << ans << endl;


    return 0;
}