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: Time Limit Exceeded)

#include <string>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
int n,h[100005];
ll dp[100005];
/*
bool compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a. != b.second){
        return a.second < b.second;
    }else{
        return a.first < b.first;
    }
}
*/
int main(){


	long long N,M;
    std::cin >> N >> M;
  
    vector<pair<long long, long long> > pairs(N);

    // 入力
    for(long long i=0; i<N; i++){
        long long a, b;
        cin >> a >> b;
        pairs[i] = make_pair(a, b);
    }

    // b順にソート
    //sort(ALL(pairs), compare_by_b);
	//sort(pairs.begin(), pairs.end(),greater<pair<int,int> >());
	sort(pairs.begin(), pairs.end());
    /*
    for(int i=0; i<N; i++){
	printf("%d",pairs[i].first);
	printf("%d\n",pairs[i].second);
	}
	*/
    long long sum;
    sum=0;
    long long ans;
    ans=0;
    long long j;
    j=0;
    for(long long i=0; i<N; i++){
		while(j<pairs[i].second){
			ans+=pairs[i].first;
			sum++;
			j++;
			if(sum==M){
				printf("%lld\n",ans);
				
			}
		}		
		j=0;
		
	}
	
}