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 int lli;

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

	vector<pair<lli,lli>> P(N);
	for (lli i = 0; i < N; i++) {
		cin >> P[i].first >> P[i].second;
	}

	sort(P.begin(), P.end());

	lli sum = 0;
	lli tmp = 0;
	lli i = 0;
	while (tmp < M) {
		sum += P[i].first * min(P[i].second, max(0LL, M - tmp));
		tmp += P[i].second;
		i++;
	}

	cout << sum << endl;

	return 0;
}