Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define rep2(x,fr,to) for(int x=(fr);x<(to);x++)
#define rep(x,to) for(int x=0;x<(to);x++)
#define repr(x,fr,to) for(int x=(fr);x>=(to);x--)
#define all(c) c.begin(),c.end()
#define sz(v) (int)v.size()

using namespace std;
typedef long long ll; typedef vector<int> VI; typedef pair<int,int> pii;
typedef vector<ll> VL; const int MD = (int)1e9 + 7;
void dbg(){ cerr << "\n"; } template <typename T,typename ...T2> void dbg(const T& fst, const T2&...rst){ cerr << fst << ": "; dbg(rst...); }
template <class T, class T2> void amax(T& a,T2 b){ if(a < b) a = b;}


int main()
{
	cin.tie(0); ios_base::sync_with_stdio(false);
	int n, m;
	cin >>n >>m;
	vector<pii> t(n);
	rep(i, n) cin >>t[i].first >>t[i].second;
	
	sort(all(t));
	ll ans = 0;
	for(auto x: t){
		ll mn = min(m, x.second);
		ans += mn * x.first;
		m -= mn;
		if(mn <= 0) break;
	}
	cout << ans <<"\n";
	return 0;
}