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: Wrong Answer)

#include <stdio.h>

typedef struct data
{
    int a,b;
} data;

void merge(data numbers[], data temp[], int left, int mid, int right)
{
    int i, left_end, num_elements, tmp_pos;

    left_end = mid - 1;
    tmp_pos = left;
    num_elements = right - left + 1; /* �z��̗v�f�� */

    /* 2�̃��X�g�ɗv�f���c���Ă��� */
    while ((left <= left_end) && (mid <= right))
    {
        if (numbers[left].a <= numbers[mid].a)
        {
            temp[tmp_pos] = numbers[left];
            tmp_pos = tmp_pos + 1;
            left = left +1;
        }
        else
        {
            temp[tmp_pos] = numbers[mid];
            tmp_pos = tmp_pos + 1;
            mid = mid + 1;
        }
    }

    /* �����̃��X�g */
    while (left <= left_end)
    {
        temp[tmp_pos] = numbers[left];
        left = left + 1;
        tmp_pos = tmp_pos + 1;
    }
    /* �E���̃��X�g */
    while (mid <= right)
    {
        temp[tmp_pos] = numbers[mid];
        mid = mid + 1;
        tmp_pos = tmp_pos + 1;
    }

    /* �����ɐ��񂷂�悤�ЂƂ̃��X�g�ɂ܂Ƃ߂� */
    for (i=0; i <= num_elements; i++)
    {
        numbers[right] = temp[right];
        right = right - 1;
    }
}

void m_sort(data numbers[], data temp[], int left, int right)
{
    int mid;

    if (right > left)
    {
        mid = (right + left) / 2; /* �z��𕪊�����ʒu */
        m_sort(numbers, temp, left, mid);
        m_sort(numbers, temp, mid+1, right);

        merge(numbers, temp, left, mid+1, right);
    }
}

void mergeSort(data numbers[], data temp[], int array_size)
{
    m_sort(numbers, temp, 0, array_size - 1);
}

int main()
{
    int n,m,i;
    long long ans = 0;
    data ab[100000],buf[100000];

    scanf("%d %d",&n,&m);
    for (i = 0; i < n; i++)
    {
        scanf("%d %d",&ab[i].a,&ab[i].b);
    }
    mergeSort(ab,buf,n);

    for (i = 0; i < n; i++)
    {
        if(m >= ab[i].b)
        {
            ans += (long long)ab[i].a * (long long)ab[i].b;
            m -= ab[i].b;
        }
        else
        {
            ans += (long long)ab[i].a * (long long)m;
            break;
        }
    }
    printf("%lld\n",ans);
}