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>

int main(){
    int n, m, i, j, work, a[100000], b[100000];
    long long int ans = 0;

    scanf("%d %d", &n, &m);

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

    for(i=0; i<n; i++){
        for(j=0; j<n; j++){
            if(a[i] < a[j]){
                work = a[i];
                a[i] = a[j];
                a[j] = work;

                work = b[i];
                b[i] = b[j];
                b[j] = work;
            }
        }
    }

    i=0;
    while(m > 0){
        if(m > b[i]){
            ans += (long long int)b[i] * (long long int)a[i];
            m -= b[i];
        }
        else{
            ans += (long long int)m * (long long int)a[i];
            m -= b[i];
        }

        i++;
    }

    printf("%lld\n", ans);

    return 0;
}