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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string>
#include <vector>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;

static const long MAX = 1e5 + 3;
int n, m, B[MAX];
long A[MAX], C[MAX][2], c, ans;

long h(long key, long i) {
    long r = (key % MAX + i * (1 + key % (MAX - 1))) % MAX;
    return r;
}

void insert(long key, long value) {
    long i = 0;

    while ( true ) {
        if ( C[h(key, i)][0] == key ) {
            break;
        } else if ( C[h(key, i)][0] ) {
            i++;
        } else {
            C[h(key, i)][0] = key;
            C[h(key, i)][1] = value;
            break;
        }
    }
}

long find(long key) {
    long i = 0;

    while ( true ) {
        if ( C[h(key, i)][0] == key ) {
            return C[h(key, i)][1];
            break;
        } else if ( C[h(key, i)][0] ) {
            i++;
        } else {
            return 0;
            break;
        }
    }
}

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

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

    c = 0;
    ans = 0;

    while ( m > 0 ) {
        m -= find(c);
        ans += c * find(c);
        c++;
    }

    ans += (c - 1) * m;
    printf("%ld\n", ans);
    return 0;
}