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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  FOR(i,0,n)
#define req(i,n) for(int i = 1;i <=n;i++)
#define pai 3.14159265358979323846
const int INF = 1001001001;
typedef long long ll;
int A[3][3], N;
bool punched[3][3];
bool ok[3][3];
using Graph = vector<vector<int>>;
vector<vector<int>> field;
vector<bool> seen;

const int MOD = 1000000007;
typedef pair<int,int> P;

//最大公約数
int gcd(int a,int b){
	if (a%b == 0){
		return b;
	}
	else{
		return gcd(b,a%b);
	}

}

//最小公倍数
int lcm(int a,int b){
	return a /gcd(a,b) * b;
}

//素数判定
bool is_prime(long long N) {
    if (N == 1) return false;
    for (long long i = 2; i * i <= N; ++i) {
        if (N % i == 0) return false;
    }
    return true;
}




int main(){
  ll n,m;
  cin >> n >> m;
  ll c[n+1],b[n+1];
  vector<pair <ll,ll>> a;
  rep (i,n){
    cin >> c[i] >> b[i];
    a.push_back({c[i],b[i]});
  }
  sort(a.begin(),a.end());
  ll ans=0,cnt=0;
  for (auto v : a){
    while (v.second>0){
      if (cnt==m) break;
      ans += v.first;
      cnt++;
      //cout << ans << endl;
      v.second--;
    }
  }
  //cout << cnt << endl;
  cout << ans << endl;

  
}