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 Python (3.4.3) (judge verdict: Accepted)

# coding: utf-8
# Your code here!

N,M = map(int,input().split(" "))
A = []
B = []

for i in range(N):
    a,b = map(int,input().split(" "))
    A.append(a)
    B.append(b)
    
L = list(zip(A,B))
L.sort()

#N 店の軒数


#M 買いたい本数
#A 一本の価格
#B 買える本数
yen = 0

while True:
    i = 0#i = A.index(min(A))
    #y = min(A)
    #print(A,B,M)
    if L[0][1] >= M:
        yen += M*L[0][0]
        break
    else:#B[i] < M
        yen += L[0][1] * L[0][0]
        M -= L[0][1]
        del L[0]
        
        
print(yen)