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 <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <complex>
#include <vector>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <algorithm>
#include <numeric>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define FOR(i, a, b) for(ll i = static_cast<ll>(a); i < static_cast<ll>(b); i++)
#define FORR(i, a, b) for(ll i = static_cast<ll>(a); i >= static_cast<ll>(b); i--)
#define REP(i, n) for(ll i = 0ll; i < static_cast<ll>(n); i++)
#define REPR(i, n) for(ll i = static_cast<ll>(n); i >= 0ll; i--)
#define ALL(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> IP;
typedef pair<ll, LP> LLP;

const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};

constexpr int INF = 100000000;
constexpr ll LINF = 10000000000000000ll;
constexpr int MOD = static_cast<int>(1e9 + 7);
constexpr double EPS = 1e-9;
// }}}

ll N, M;
vector<LP> v;

void init()
{
}

void solve()
{
    sort(ALL(v));
    ll ans = 0;
    REP(i, N){
        int b = min(M, v[i].se);
        ans += v[i].fi * b;
        M -= b;
        if(M <= 0) break;
    }
    cout << ans << endl;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N >> M;
    v.resize(N);
    REP(i, N) cin >> v[i].fi >> v[i].se;
    solve();
    return 0;
}

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