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>
#define fi first
#define se second

using namespace std;
using pii=pair<int,int>;
const int mxn = 1e5+10;
pii arr[mxn];

int main() {
    int n, m;
    cin >> n >> m;
    for(int i=0; i<n; i++) {
        cin >> arr[i].fi >> arr[i].se;
    }
    sort(arr, arr+n);
    long long sum=0;
    for(int i=0; (i<n) && m; i++) {
        for(; arr[i].se-- && m;) sum+=arr[i].fi, m--;
    } cout << sum;
    return 0;
}