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;
using P = pair<int,int>;
using Pl = pair<ll,ll>;
#define REP(i,n) for(ll i = 0;i < (ll)n;i++)
#define REPD(i,n) for(ll i = (ll)n - 1;i >= 0;i--)
#define ALL(x) (x).begin(),(x).end()
#define FILLl(n,first_dimension_size,value) fill((ll*)n,(ll*)(n+first_dimension_size),value)
#define FILL(n,first_dimension_size,value) fill((int*)n,(int*)(n+first_dimension_size),value)
#define INF (1 << 29)
#define MOD 1000000007
//int dx[4] = {1,0,0,-1};
//int dy[4] = {0,1,-1,0};



int main(){

  int n,m;
  cin >> n >> m;
  ll res = 0,acc = 0;
  vector<Pl> v;
  REP(i,n){
    ll x,y;cin >> x >> y;
    v.push_back(make_pair(x,y));
  }
  sort(ALL(v));
  for(ll i = 0;i < n;i++){
    for(ll j = 0;j < v[i].second;j++){
      if(acc >= m)break;
      acc++;
      res += v[i].first;
    }
  }
  cout << res << endl;

  return 0;
}