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 <iostream>
#include <vector>
#include <utility>
#include <algorithm>

using namespace std;

int main() {
	long long N, M;
	cin >> N >> M;

	vector<pair<long long, long long> > s(N);
	for(auto & x: s) cin >> x.first >> x.second;

	sort(s.begin(), s.end());
	long long out = 0;
	for(int i = 0; M > 0; i++) {
		out += s[i].first * min(s[i].second , M);
		M -= min(s[i].second, M);
	}

	cout << out << endl;
}