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;
using ll = long long;
#define rep(i,n) for (ll i = 0; i< (n); ++i)
const long long INF = 1LL<<58;

// aを常に小さい値にする
template<class T> void chmin(T &a, T b){ if(a>b) a = b; }

// 二分探索 前方のindexを返すもの（自分自身を含む） 前方: -INF側のこと
template<class T> int former(const vector<T> &v, T x){
    return upper_bound(v.begin(), v.end(), x) - v.begin() -1; // upper_bound : keyより大きい要素の内の一番左側のイテレータを返す
}
// 二分探索 後方のindexを返すもの（自分自身を含む）後方: INF側のこと
template<class T> int latter(const vector<T> &v, T x){
    return lower_bound(v.begin(), v.end(), x) - v.begin(); // lower_bound : key以上の要素の内の一番左側のイテレータを返す
}

/*
 *
 *
 *
 * */

int main() {

    int N;
    ll M;
    cin >> N >> M;
    vector<pair<ll, ll>> v(N);
    for (int i = 0; i < N; ++i)
        cin >> v[i].first >> v[i].second;
    sort(v.begin(), v.end());

    ll res = 0;
    for (int i = 0; i < N && M > 0; ++i) {
        res += v[i].first * min(M, v[i].second);
        M -= min(M, v[i].second);
    }
    cout << res << endl;
}