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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>

using namespace std;

#define FOR(i,a,b) for(int i = (a); i < (b); ++i)
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()

inline int toInt(string s) {int v; istringstream sin(s);sin>>v; return v;}

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector< vector<int> > Mat;
typedef tuple<int, int, int, int> T;

const int MOD = (int)1e9+7;
const int INF = (int)1e9;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int ddx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[] = {1, 1, 0, -1, -1, -1, 0, 1};

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

    ll N, M;
    cin >> N >> M;
    vector< pair<ll, ll> > drink(N);
    rep(i, N) cin >> drink[i].first >> drink[i].second;

    sort(drink.begin(), drink.end());

    ll ans = 0;
    rep(i, N){
        if(drink[i].second <= M){
            ans += drink[i].first * drink[i].second;
            M -= drink[i].second;
        }else{
            ans += drink[i].first * M;
            break;
        }
    }

    cout << ans << endl;

    return 0;
}