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(void){
    int n = 0;
    int m = 0;
    int t = 0;
    int c = 0;
    scanf("%d %d", &n, &m);
    long int a[n], b[n];
    for(int i=0;i<n;i++){
        scanf("%ld %ld", &a[i], &b[i]);
    }

    for(int i=0;i<n-1;i++){
        for(int j=0;j<n-1;j++){
            if (a[j+1]<a[j]){
            //a
            t = a[j];
            a[j] = a[j+1];
            a[j+1] = t;
            //b
            t = b[j];
            b[j] = b[j+1];
            b[j+1] = t;
            }
        }
    }
    long int sum = 0;
    for(int i=0;i<n;i++){
        if(m ==0)break;
        long int d = b[i];
        for(int j=0;j<b[i];j++){
            if(m == 0)break;
            sum += a[i];
            d -= 1;
            m -= 1;
            //printf("%d %d %d %d\n", a[i], d, sum, m);
            if(m == 0)break;

        }
        if(m == 0)break;
    }
    printf("%ld\n", sum);
}