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 <iostream>
#include <vector>
#include<list>
#include <algorithm>
#include <string>
#include <math.h>
#include<map>
#include <utility>

#include<queue>
using namespace std;
#define REP(i,n) for(int i=0;i<n;i++)
 
#define INF (1e9)
typedef long long unsigned int ll;
#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))
 

 
int vector_finder(std::vector<int> vec, int number) {
  size_t index = std::distance( vec.begin(), std::find(vec.begin(), vec.end(), number) );
  if (index != vec.size()) { // 発見できたとき
    return 1;
  }
  else { // 発見できなかったとき
    return 0;
  }
}


int main()
{
	int N, M;
	cin >> N >> M;
	vector<pair<int, int> > mp;
	int A[N]; 
	int B[N];
	REP(i,N){
		cin >> A[i] >> B[i];
		pair<int, int> p = make_pair(A[i], B[i]);
		mp.push_back(p);
	}

	sort(mp.begin(), mp.end());

	ll ans = 0;
	pair<int, int> current = mp[0];
	int idx = 0;
	REP(i, M)
	{
		if(current.second==0){
			current = mp[idx + 1];
			idx++;
		}
		ans += current.first;
		current.second--;
	}
	cout << ans << endl;

	return 0;
}