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: Accepted)

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

#define ULL unsigned long long

int cmp(const void *a, const void *b) {
    int *p = *(int**)a; int *q = *(int**)b;
    if( p[0] > q[0] )       { return 1; }
    else if( p[0] < q[0] )  { return -1; }
    else                    { return 0; }
}

int main()
{
    int i, n, m, temp;
    ULL mny;
    scanf("%d %d", &n, &m);
    int **a = (int **)malloc(n * sizeof(int *));
    a[0] = (int *)malloc(n*2*sizeof(int));
    for(i = 1; i < n; ++i) { a[i] = a[0] + i*2; }
    for(i = 0; i < n; ++i) { scanf("%d %d", &a[i][0], &a[i][1]); }
    qsort(a, n, sizeof(int*), cmp);
    //for(i = 0; i < n; ++i) { printf("%d %d\n", a[i][0], a[i][1]); }

    //printf("---------\n");
    i = 0; mny = (ULL)0;
    for(i = 0; (i < n && m > 0); ++i) {
        temp = ( a[i][1] <= m) ? a[i][1] : m;
        mny += (ULL)a[i][0] * (ULL)temp;
        m -= temp;
    }
    printf("%llu\n", mny);
    return 0;
}