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>
#define rep(i, n) for (ll i = 0; i < (ll)(n); ++i)
#define rep2(i, s, n) for (ll i = s; i < (ll)(n); i++)
#define repr(i, n) for (ll i = n; i >= 0; i--)
#define pb push_back
#define COUT(x) cout << (x) << endl
#define COUTF(x) cout << setprecision(15) << (x) << endl
#define ENDL cout << endl
#define DF(x) x.erase(x.begin())  // 先頭文字削除
#define ALL(x) x.begin(), x.end()
#define SORT(x) sort(ALL(x))
#define REVERSE(x) reverse(ALL(x))
#ifdef _DEBUG
#define debug(x) cout << "[debug] " << #x << ": " << x << endl
#else
#define debug(x)
#endif
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using P = pair<ll, ll>;
constexpr ll INF = 0x3f3f3f3f3f3f3f3f;
constexpr double PI = 3.141592653589793238462643383279;
ll getDigit(ll x) {
  return x == 0 ? 1 : log10(x) + 1;
}

ll gcd(ll x, ll y) {
  return y ? gcd(y, x % y) : x;
}

ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

vector<P> factorize(ll n) {
  vector<P> result;
  for (ll i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      result.pb({i, 0});
      while (n % i == 0) {
        n /= i;
        result.back().second++;
      }
    }
  }
  if (n != 1) {
    result.pb({n, 1});
  }
  return result;
}

template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

template <typename T>
void debugV(const vector<T> v) {
#ifdef _DEBUG
  rep(i, v.size()) {
    cout << i << ":" << v[i] << " ";
  }
  cout << endl;
#else
  (void)v;
#endif
}

signed main() {
  ll N, M;
  cin >> N >> M;

  vector<P> p;
  rep(i, N) {
    ll a, b;
    cin >> a >> b;
    p.pb({a, b});
  }

  SORT(p);

  ll ans = 0;
  ll count = 0;
  rep(i, N) {
    P pp = p[i];
    rep(j, pp.second) {
      ans += pp.first;
      count++;
      if (count == M) {
        COUT(ans);
        return 0;
      }
    }
  }

  return 0;
}