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 <algorithm>
#include <iomanip>
#include <iostream>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define SORT(a) sort((a).begin(), (a).end())
#define RSORT(a) sort((a).rbegin(), (a).rend())
#define REP(i, n) for (int i = 0; i < n; i++)
#define RREP(i, n) for (int i = n; 0 <= i; i--)
#define FOR(i, start, end) for (int i = start; i < end; i++)
#define FORSTEP(i, start, end, step) for (int i = start; i < end; i += step)
#define ALL(a) a.begin(), a.end()
typedef long long ll;
const int INF = 100100100;
using namespace std;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N, M;
    cin >> N >> M;
    vector<pair<int, int>> A(N);
    REP(i, N) {
        int a, b;
        cin >> a >> b;
        A[i].first  = a;
        A[i].second = b;
    };
    SORT(A);

    ll sum = 0;
    ll cnt = 0;
    REP(i, N) {
        cnt += A[i].second;
        sum += (ll)A[i].first * A[i].second;
        if (cnt >= M) {
            ll diff = cnt - M;
            sum -= (ll)A[i].first * diff;
            break;
        }
    }

    cout << sum << endl;

    return 0;
}