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 FOR(i,n) for(int i=0; i<n; ++i)
#define FORR(i,a,n) for(int i=a; i<n; ++i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
 
int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }
void printfv(VI a){ int n = a.size(); FOR(i, n) cout << a[i] << " "; }

int main(void){

    int n, m;
    cin >> n >> m;

    vector<PL> vv;
    ll price, num, num_total = 0, amount = 0;
    FOR(i, n){
        cin >> price >> num;
        vv.push_back(make_pair(price, num));
    }

    sort(ALL(vv));
    FOR(i, n){
        //cout << vv[i].first << " " << vv[i].second << endl;

        if((num_total+vv[i].second) > m){
            amount += (m - num_total)*vv[i].first;
            break;
        } else {
            amount += vv[i].second*vv[i].first;
        }

        //cout << amount << endl;
        num_total += vv[i].second;
    }

    cout << amount;
    
    return 0;
}