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: Runtime Error)

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <list>
#include <queue>
#include <deque>
#include <algorithm>
#include <numeric>
#include <utility>
#include <complex>
#include <functional>
using namespace std;
typedef long long ll;
const int MAX_N = 10 ^ 5;
int N, M;
//ll a[MAX_N];
//ll b[MAX_N];
vector<pair<ll, ll> > p(MAX_N);
void solve(){
    sort(p.begin(),p.end());
    int i=-1;
    ll ans=0;
    while(M>0){
        i++;
        ans+=p[i].first*p[i].second;
        M-=p[i].second;
    }
    ans+=(ll)M*p[i].first;
    cout<<ans<<'\n';
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N >> M;
    for (int i = 0; i < N; i++){
        cin>>p[i].first>>p[i].second;
    }
    solve();
}