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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pa;
#define MM 1000000000
#define MOD MM+7
#define MAX 10100000
#define INF 1e15
#define MAP 110
#define initial_value -1
#define LINF 1LL << 60
#define Pair pair<int,int>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
int dx[4] = {-1,0,1,0};
int dy[4] = {0,-1,0,1};
ll A[100100],B[100100];
vector<pa> PP;
int main(){
    ll N,M; cin >> N >> M;
    for(ll i = 0; i < N; i++){
        cin >> A[i] >> B[i];
        PP.push_back(Pair(A[i],B[i]));
    }
    sort(PP.begin(),PP.end());
    ll m = M;
    ll ans = 0;
    for(ll i = 0; i < N; i++){
        if(A[i] > m){
            m -= A[i];
            ans += A[i]*B[i];
        } else {
            ans += m*B[i];
            break;
        }
    }
    cout << ans << endl;
}