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;
const int SZ = 100005;
typedef long long ll;
pair<ll,ll> arr[SZ];
int main(void){
	ll n,m; scanf("%lld %lld",&n,&m);
	for(int i=0;i<(int)n;i++){
		ll x,y; scanf("%lld %lld",&x,&y);
		arr[i]=make_pair(x,y);
	}
	sort(arr,arr+n);
	long long ans = 0LL;
	int p = 0;
	while(m){
		ll buy = min(m,arr[p].second);
		ans += buy*arr[p].first;
		p++;
		m-=buy;
	}
	printf("%lld",ans);
	return 0;
}