A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.

Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.

You want to buy at least L liters of lemonade. How many roubles do you have to spend?

Input

The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.

Output

Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.

Examples

Input

4 12
20 30 70 90


Output

150


Input

4 3
10000 1000 100 10


Output

10


Input

4 3
10 100 1000 10000


Output

30


Input

5 787787787
123456789 234567890 345678901 456789012 987654321


Output

44981600785557577

Note

In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.

In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.

In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, L, cost = 0;
  cin >> n >> L;
  vector<pair<long long, long long> > bot(n);
  for (int i = 0; i < n; i++) {
    cin >> bot[i].first;
    bot[i].second = (1LL << i);
  }
  for (int i = 1; i < n; i++) {
    if (bot[i].first >= bot[i - 1].first * 2) {
      bot[i].first = bot[i - 1].first * 2;
    }
  }
  long long tl = L;
  while (L > 0) {
    long long p, c;
    p = L;
    c = LONG_LONG_MAX;
    for (int i = n - 1; i >= 0; i--) {
      if (bot[i].second >= L) {
        if (bot[i].first < c) {
          p = bot[i].second;
          c = bot[i].first;
        }
      } else {
        long long req = ceil((L + 0.0) / bot[i].second);
        long long xcost = bot[i].first * req;
        if (xcost < c) {
          c = xcost;
          p = (req * bot[i].second);
        }
      }
    }
    cost += c;
    L -= p;
  }
  L = tl;
  long long last = 0, first = 0;
  for (int i = 0; i < 32; i++)
    if (L & (1LL << i)) last = i;
  for (int i = 0; i < 32; i++)
    if (L & (1LL << i)) {
      first = i;
      break;
    }
  long long maxL = (1LL << (last + 1));
  if (1) {
    while (L < maxL) {
      long long c = 0;
      for (int i = 0; i < 30; i++) {
        if (L & (1LL << i)) {
          if (i < n)
            c = (c + bot[i].first);
          else {
            long long z = ceil(((1LL << i) + 0.0) / bot.back().second);
            c = c + (z * bot.back().first);
          }
        }
      }
      cost = min(cost, c);
      L = L + (L & (-L));
    }
  }
  cout << cost;
}
