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<math.h>
#include<stdlib.h>
#include<string.h>
#include<stdbool.h>
#include<limits.h>


#define MOD 998244353

#define MIN(a,b) ((a)<(b)?(a):(b));
#define MAX(a,b) ((a)>(b)?(a):(b));

typedef long long int64;

int main(void){
    int N,M;

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

    int A[N],B[N];

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

    int num=0,temp=INT_MAX,I;

    long long ans;

    while(num<M){
        I=0;
        for(int i=0;i<N;i++){
            temp=MIN(temp,A[i]);
            if(temp==A[i]){
                I=i;
            }
        }
        if(num+B[I]>M){
            ans+=(long long)A[I]*(B[I]-(num+B[I]-M));
            // printf("%d %d\n",A[I],(B[I]-(num+B[I]-M)));
            // printf("%lld\n",ans);
        }else{
            ans+=(long long)A[I]*B[I];
            // printf("%d %d\n",A[I],B[I]);
            // printf("%lld\n",ans);
        }
        A[I]=INT_MAX;
        temp=INT_MAX;
        num+=B[I];
    }

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