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>


//二次元配列quicksort 参考http://www.math.u-ryukyu.ac.jp/~suga/C/2004/13/node6.html
//http://d.hatena.ne.jp/lagos_on/20090517/1242586942
void qsort(int list1[],int list2[], int left, int right) {
    int l = left;
    int r = right;
    int pivot = list1[(left + right) / 2]; // 中央からピボットを選択
    int temp;

    while (1) {
        while (list1[l] < pivot) l++;
        while (list1[r] > pivot) r--;

        if (l > r) break;

        // pivotより小さい値を左、大きい値を右に入れ替え
        temp = list1[l];
        list1[l] = list1[r];
        list1[r] = temp;
        
        temp = list2[l];
        list2[l] = list2[r];
        list2[r] = temp;


        l++, r--;
    };

    // pivotより小さい領域と大きい領域をそれぞれ再帰的にソート
    if (left < r)  qsort(list1,list2 ,left, r);
    if (l < right) qsort(list1,list2, l, right);
}

void quicksort(long long int list[],int list2[], int left, int right)
{
    int i, last;
    int temp;
    if (left >= right)
        return;

    last = left;
    for (i=left+1; i <= right; i++){
        if (list[i] < list[left] ){
            last++;
            temp=list[last];
            list[last]=list[i];
            list[i]=temp;
            temp=list2[last];
            list2[last]=list2[i];
            list2[i]=temp;
        }
    }

    temp=list[left];
    list[left]=list[last];
    list[last]=temp;

    temp=list2[left];
    list2[left]=list2[last];
    list2[last]=temp;

    quicksort(list,list2, left, last-1);
    quicksort(list,list2, last+1, right);
}
int main(void){
    int N,M;
    scanf("%d %d",&N,&M);
    int A[N];
    int B[N];
    for(int i=0;i<N;i++){
        scanf("%d %d",&A[i],&B[i]);
    }
    long long int money = 0;
    qsort(A,B,0,N-1);

    printf("------------\n");
    for(int i=0;i<N;i++){
        printf("%d %d\n",A[i],B[i]);
    }
//*/
    for(int i=0;i<N;i++){
        if(M>B[i]){
            money += (long long int)A[i]*B[i];
            M -= B[i];
            //printf("money = %lld ,M=%d\n",money,M);
        }else{
            money += (long long int)A[i]*M;
            M -= M;
            break;
        }
    }

    printf("%lld",money);
    return 0;
}