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: Accepted)

#include <stdio.h>

typedef	unsigned long long ull;

ull temp[100000];    /* 最小でも配列と同じサイズの領域が必要 */
ull op_temp[100000];    /* 最小でも配列と同じサイズの領域が必要 */

void MergeSort(ull x[], ull op[], ull left, ull right)
{
    ull mid, i, j, k;
	
    if (left >= right)              /* 配列の要素がひとつなら */
        return;                     /* 何もしないで戻る */

                                    /* ここでは分割しているだけ */
    mid = (left + right) / 2;       /* 中央の値より */
    MergeSort(x, op, left, mid);        /* 左を再帰呼び出し */
    MergeSort(x, op, mid + 1, right);   /* 右を再帰呼び出し */

      /* x[left] から x[mid] を作業領域にコピー */
    for (i = left; i <= mid; i++){
        temp[i] = x[i];
		op_temp[i] = op[i];
	}
      /* x[mid + 1] から x[right] は逆順にコピー */
    for (i = mid + 1, j = right; i <= right; i++, j--){
        temp[i] = x[j];
		op_temp[i] = op[j];
	}

    i = left;         /* i とj は作業領域のデーターを */
    j = right;        /* k は配列の要素を指している */

    for (k = left; k <= right; k++)    /* 小さい方から配列に戻す */
        if (temp[i] <= temp[j]){        /* ここでソートされる */
			op[k] = op_temp[i];
            x[k] = temp[i++];
		}
        else{
			op[k] = op_temp[j];
            x[k] = temp[j--];
		}
}

int main(void)
{
	ull n, m;
	ull a[100000], b[100000];
	ull buy_m=0, sum_money=0;
	ull i, j;
	
	scanf("%lld %lld", &n, &m);
	for(i=0; i<n; i++)
		scanf("%lld %lld", &a[i], &b[i]);
	
	MergeSort(a, b, 0,n-1);
	//for(i=0; i<n; i++)
	//	printf("%lld %lld\n", a[i], b[i]);

	for(i=0; i<n; i++){
		if(buy_m+b[i] < m){
			buy_m += b[i];
			sum_money += a[i]*b[i];
		}
		else{
			ull diff_m = m-buy_m;
			sum_money += a[i]*diff_m;
			break;
		}
	}

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

	return 0;
}