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
using ll = int64_t;
using ul = uint64_t;
using ld = long double;
const ll MOD = 1e9 + 7;
const ll INF = LONG_MAX;
const ul MAX = ULONG_MAX;
const char nl = '\n';

#define elif else if
#define def inline auto
#define run __attribute__((constructor)) def _

// Debug
#define debug(x) cerr << (x) << " (L:" << __LINE__ << ")" << '\n'

// Loop
#define inc(i, a, n) for (ll i = (a), _##i = (n); i < _##i; ++i)
#define dec(i, a, n) for (ll i = (a), _##i = (n); i > _##i; --i)
#define rep(i, n) inc(i, 0, n)
#define be(c) begin(c), end(c)

// Stream
#define fout(n) cout << fixed << setprecision(n)
#define fasten cin.tie(0), ios::sync_with_stdio(0)

// Speed
run() { fasten, fout(10); }
#pragma GCC optimize("-O3")
#pragma GCC target("avx")

signed main() {
    int n, m;
    cin >> n >> m;
    using pii = pair<int, int>;
    vector<pii> pis(n);
    for(auto& e: pis) {
        cin >> e.first >> e.second;
    }
    sort(be(pis));
    ll ans = 0;
    int bought = 0;
    rep(i, n) {
        auto& e = pis[i];
        auto b = min(m - bought, e.second);
        ans += (ll)e.first*b;
        bought += b;
        if (bought == m) break;
    }
    cout << ans << nl;
}