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;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep2(i, a, n) for (int i = (a); i < (n); i++)
#define repb(i, n) for (int i = (n - 1); i >= 0; i--)
#define repb2(i, a, n) for (int i = (n - 1); i >= (a); i--)
#define all(x) x.begin(), x.end()
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
inline void fastio() { ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); }

int main() {
    fastio();

    int N, M;
    vector<pair<ll, ll>> X;
    cin >> N >> M;
    rep(i, N) {
        int a, b;
        cin >> a >> b;
        X.push_back({a, b});
    }
    sort(all(X));

    ll ans = 0;
    for (auto &x : X) {
        int price = x.first;
        int bottles = x.second;
        ll t = (bottles <= M) ? bottles : M; // 今回の購入本数
        ans += t * price;
        M -= t; // 残りの購入本数
        if (!M) {
            break;
        }
    }
    co(ans);
}