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;
typedef long long ll;
ll n,m;
const int max_n = 1e6;
pair<ll, ll> p[max_n];
void solve(){
  ll sum = 0;
  for(int i=0 ; i<n ; i++){
    ll k = min(p[i].second, m);
    sum += k * p[i].first;
    m -= k;
    if(m == 0) break;
  }
  cout << sum << endl;
}
int main(){
  cin >> n >> m;
  for(int i=0 ; i<n ; i++){
    cin >> p[i].first >> p[i].second;
  }
  sort(p, p+n);
  solve();
  return 0;
}