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>

#ifdef DEBUG
#define PRINT(x)\
    cout<<"func "<<__func__<<": line "<<__LINE__<<": "<<#x<<" = "<<(x)<<endl;
#define PRINTA(a,first,last)\
    cout<<"func "<<__func__<<": line "<<__LINE__<<": "<<#a<<"["<<(first)<<", "<<(last)<<")"<<endl;\
    for (int i=(first);i<(last);++i){cout<<#a<<"["<<i<<"] = "<<(a)[i]<<endl;}
#else
#define PRINT(x)
#define PRINTA(a,first,last)
#endif

#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()
#define MOD 1000000007

using namespace std;

using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;

template <class T, class U> void amax(T& x, U y) {if (x < y) x = y;}
template <class T, class U> void amin(T& x, U y) {if (x > y) x = y;}

int main(void)
{
    ll N, M;
    cin >> N >> M;
    using P = pair<ll, ll>;
    vector<P> AB(N);
    REP(i, N) {
        ll A, B;
        cin >> A >> B;
        AB[i] = P(A, B);
    }
    sort(all(AB));
    ll result = 0;
    REP(i, N) {
        ll A = AB[i].first;
        ll B = AB[i].second;
        PRINT(A);
        PRINT(B);
        int x = min(M, B);
        PRINT(x);
        result += A * x;
        M -= x;
        if (M == 0) {
            break;
        }
    }
    cout << result << endl;

    return 0;
}