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>

int main(){

    int i, j, N, M, ans=0, katta=0, breakf=0;

    scanf("%d %d", &N, &M);

    int pri[N], zai[N];

    for (i=0; i<N; i++){
        scanf("%d %d", &pri[i], &zai[i]);
    }

    int sequence[N], max, max_num, sude[N];
    for (i=0; i<N; i++){
        sude[i] = 0;
    }

    for (i=0; i<N; i++){
        max=0;
        for(j=0; j<N; j++){
            if ((pri[j] > max) && sude[j] == 0){
                max = pri[j];
                max_num = j;
            }
        }
        sequence[i] = max_num;
        sude[max_num] = 1;
    }


    for(i=0; i<N; i++){
        for(j=0; j<zai[sequence[N-i-1]]; j++){
            if (katta == M){
                breakf=1;
                break;
            }
            ans+=pri[sequence[N-i-1]];
            katta++;
        }
        if (breakf==1) break;
    }


    printf("%d", ans);

    return 0;
}