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: Wrong Answer)

#include <bits/stdc++.h>
 
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,n) for(int i=1;i<=(n);i++)
#define lol long long
#define SUM(n) ((n)+1)*(n)/2　//1〜nまでの総和を求める式
#define mp make_pair
#define fi first
#define se second
#define pu push_back
#define SYOU(x) fixed<<setprecision(x+1) //小数点桁数を指定する
#define abs(x,y) max(x,y)-min(x,y)
#define all(v) v.begin(),v.end()
#define UPDight(a,b) (a+b-1)/b //小数点切り上げ 
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const int MOD=int(1e9)+7; 
using namespace std;
using pii = pair<int,int>;
typedef vector<int> vit;

signed main(void){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);

  int n, m;
  cin >> n >> m;
  priority_queue<pii, vector<pii>,greater<pii> > ab;
  int a, b;
  rep(i, n){
    cin >> a >> b;
    ab.emplace(a, b);
  }
  lol ans = 0;
  while(m > 0){
    a = ab.top().fi;
    b = ab.top().se;
    ab.pop();
    ans += a * min(m, b);
    m -= min(m, b);
    //  cout << a << " " << b << '\n';
  }
  cout << ans << '\n';
  return 0;
}