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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

type
  pair = record
    a : int64;
    b : integer;
  end;
	pairArray = array[1..100000] of pair;
var i,n,m:integer;
    input:pairArray;
    sum : int64;

procedure	swap(var a,b:pair);
var tmp:pair;
begin
	tmp:=a;
	a:=b;
	b:=tmp
end;

procedure QuickSort(var a:pairArray;start_index,end_index:integer);
var i,j,x:int64;
begin
	x:=a[(start_index+end_index)div 2].a;
	i:=start_index;
	j:=end_index;
	while(true)do
	begin
		while(a[i].a<x)do i:=i+1;
		while(x<a[j].a)do j:=j-1;
		if(i<j)then
		begin
			swap(a[i],a[j]);
			i:=i+1;
			j:=j-1;
		end else
		begin
			if(start_index<i-1)then QuickSort(a,start_index,i-1);
			if(j+1<end_index)then QuickSort(a,j+1,end_index);
			exit;
		end;
	end;
end;

begin
  sum := 0;
  readln(n,m);
  for i := 1 to n do read(input[i].a,input[i].b);
  QuickSort(input,1,n);
  for i := 1 to n do begin
    if m > input[i].b then begin
      sum := sum + input[i].b * input[i].a;
      m := m - input[i].b
    end else begin
      sum := sum + m * input[i].a;
      m := 0
    end;
    if m = 0 then break;
	end;
  writeln(sum)
end.