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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
int main(void){
    unsigned long n, m, a[1000], b[1000], cost, min, smin, i, j;
    scanf("%ld %ld", &n, &m);
    for(i=0, j=0; i<n; i++){
        scanf("%ld %ld", &a[i], &b[i]);
        if(i==0)
            min = a[i];
        if(a[i]<min){
            min = a[i];
            j = i;
        }
    }
    if(b[j]<m)
        cost = min * b[j];
    else
        cost = min * m;
    while(b[j]<m){
        m -= b[j];
        for(i=0, j=0; i<n; i++){
            if(i==0)
                smin = a[i];
            if(a[i]>min && a[i]<smin){
                smin = a[i];
                j = i;
            }
        }
        min = smin;
        if(b[j]<m)
            cost += min * b[j];
        else
            cost += min * m;
    }
    printf("%ld\n", cost);
    
    return 0;
}