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: Accepted)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define pb push_back
#define mp make_pair

// random generator
// mt19937 rndm(time(0));

// ll gcd(ll a, ll b)
// {
//     while (b>0)
//     {
//         ll tmp = a;
//         a = b;
//         b = tmp%b;
//     }
//     return a;
// }


// long long gcd(long long a, long long b) {
//  return a? gcd(b % a, a) : b;
// }


const int M=1e9+7;
ll powmod(ll a,ll b, ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res%mod;}/* a^b%mod */
const ll N=1e5+10;


int main() {
    
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    
    int n, m;
    cin>>n>>m;
    vector<pair<ll, ll> >v;
    for(int i=0;i<n;++i) {
        int a, b;
        cin>>a>>b;
        v.pb(mp(a, b));
    }
    sort(v.begin(), v.end());
    ll ans = 0;
    for(int i=0;i<n;++i) {
        if(m<=0)break;
        if(m>v[i].second) {
            m-=v[i].second;
            ans+=v[i].second*v[i].first;
        }
        else {
            ans+=m*v[i].first;
            m=0;
        }
    }
    cout<<ans<<endl;
    return 0;
}