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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

void BubSort(long x[],long y[], int n)
{
    int i, j, temp;

    for (i = 0; i < n - 1; i++)
    {
        for (j = n - 1; j > i; j--)
        {
            if (x[j - 1] > x[j])
            {
                temp = x[j];
                x[j] = x[j - 1];
                x[j - 1] = temp;

                temp = y[j];
                y[j] = y[j - 1];
                y[j -1] = temp;

            }
        }
    }
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    long a[n], b[n];
    int i;
    long long ans=0;
    for(i = 0; i < n; i++){
        scanf("%ld%ld",&a[i], &b[i]);
    }
    BubSort(a,b,n);
    for(i=0; i<n; i++){
        if(m >= b[i]){
            ans += a[i]*b[i];
            m -= b[i];
        }else{
            ans += m*a[i];
            break;
        }
    }
    printf("%lld\n",ans);

    return 0;
}