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 <bits/stdc++.h>

typedef long long int ll;
typedef std::pair<ll, int> pii;
using namespace std;

bool s(pii a, pii b) {
    return a.first <= b.first;
}

int main() {
    int N, M;
    vector<pii> v;
    cin >> N >> M;
    ll ans = 0;
    for(int i=0;i<N;i++) {
        int a,b;
        cin >> a >> b;
        v.push_back(make_pair(a,b));
    }
    sort(v.begin(), v.end(), s);
    int c=0;
    for(auto i: v) {
        ans+=(ll)(i.first * i.second);
        c+=i.second;
        if(c>M) {
            ans -= (ll)(i.first*(c-M));
            cout << ans;
            return 0;
        } else if(c==M) {
            cout << ans;
            return 0;
        }
    }
}