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 swap(int* array, int i, int j) {
    int temp;
    temp = array[i];
    array[i] = array[j];
    array[j] = temp;
}
int max(int x, int y) {
    if (x > y) {
        return x;
    } else {
        return y;
    }
}
int min(int x, int y) {
    if (x > y) {
        return y;
    } else {
        return x;
    }
}
int main(void) {
    unsigned long long int n, m, a[200000], b[200000];
    int i, j, k;
    unsigned long long int ans = 0, temp;
    int sum = 0;
    scanf("%llu", &n);
    scanf("%llu", &m);
    for (i = 0; i < n; i++) {
        scanf("%llu %llu", &a[i], &b[i]);
    }

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

    sum = 0;
    ans = 0;
    for (i = 0; i < n && sum < m; i++) {
        if (m - sum > b[i]) {
            sum += b[i];
            // printf("sum==%d\n", sum);
            ans += a[i] * b[i];
        } else {
            // printf("checked sum==%d\n", sum);
            ans += a[i] * (m - sum);
            sum += m - sum;
        }
    }
    printf("%llu\n", ans);
    return 0;
}