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>

void b_sort(int n, int *a, int *b)
{
    int i, j;
    int temp;

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

int main(void)
{
    int n, m;
    int a[100000];
    int b[100000];
    int at, bt;

    int i, j, n_unique;
    int bin;
    long long int total;
    int a_sorted[100000];
    int b_sorted[100000];

    n_unique = 0;
    scanf("%d %d", &n, &m);
    for(i=0;i<n;i++){
        scanf("%d %d", &a[i], &b[i]);
    }
    b_sort(n, a, b);

    bin = 0;
    total = 0;
    for(i=0;i<n;i++){
        if(bin + b[i] < m){
            total += (long long int)a[i] * (long long int)b[i];
            bin += b[i];
        }else{
            total += (long long int)(m - bin) * (long long int)a[i];
            bin += (m - bin);
            break;
        }
    }

    printf("%lld", total);

    return 0;
}