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>
#include <stdlib.h>

int main(void)
{
    int N, M;
    int *A, *B, *R, *R2;
    char c;
    
    int i, j;
    int Ai, Bi;
    long long int cost;
    int cans;
    
    
    scanf("%d%c%d", &N, &c, &M);
    
    
    A = (int*)malloc(sizeof(int)*N);
    B = (int*)malloc(sizeof(int)*N);
    R = (int*)malloc(sizeof(int)*N);
    R2 = (int*)malloc(sizeof(int)*N);
    
    
    scanf("%d%c%d", &A[0], &c, &B[0]);
    
    R[0] = 0;
    
    for(i=1; i<N; i++)
    {
        scanf("%d%c%d", &A[i], &c, &B[i]);
        
        R[i] = i;
        
        for(j=0; j<i; j++)
        {
            if(A[i] <= A[j])
            {
                if(R[i] > R[j])
                    R[i] = R[j];
                    
                R[j] = R[j]+1;
            }
        }
    }
    
    
    for(i=0; i<N; i++)
    {
        R2[R[i]] = i;
    }
    
    
    cost = 0;
    cans = M;
    
    for(i=0; i<N; i++)
    {
        if(B[R2[i]] < cans)
        {
            cost += (long long int)A[R2[i]]*B[R2[i]];
            cans -= B[R2[i]];
        }
        else
        {
            cost += (long long int)A[R2[i]]*cans;
                    
            break;
        }
    }
    
    
    printf("%lld\n", cost);
    
    
    free(A);
    free(B);
    free(R);
    free(R2);
    
    
    //system("pause");
    
    
    return 0;
}