Problem: 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.

Examples

Input

2 5
4 9
2 4


Output

12


Input

4 30
6 18
2 5
3 10
7 9


Output

130


Input

1 100000
1000000000 100000


Output

100000000000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main () {
  
  int N, M;
  cin >> N >> M;
  
  vector<pair<int, int>> P(N);
  int A, B;
  for (int i = 0; i < N; i++) {
    cin >> A >> B;
    P[i] = make_pair(A, B);
  }
  
  sort(P.begin(), P.end());
  
  long long ans = 0;
  
  for (int i = 0; i < N; i++) {
    int buy = min(M, P[i].second);
    ans += P[i].first * (long long)buy;
    M -= buy;
  }
  
  cout << ans << endl;
  
}