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;
long long n;
long long l;
long long c[100];
long long cost[100];
long long num[31];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> l;
  for (int i = 0; i < n; i++) cin >> c[i];
  cost[0] = c[0];
  for (int i = 1; i < n; i++) cost[i] = min(2 * cost[i - 1], c[i]);
  num[0] = 1;
  for (int i = 1; i < 31; i++) num[i] = num[i - 1] * 2;
  long long x = l;
  long long ans = 0;
  long long anssofar = -1;
  while (x > 0) {
    long long y = log2(x);
    y = min(n - 1, y);
    long long p = x / num[y];
    x -= p * num[y];
    ans += p * cost[y];
    long long z = -1;
    for (int i = 0; i < n; i++) {
      if (x > num[i]) continue;
      if (z == -1) z = cost[i];
      z = min(cost[i], z);
    }
    z += ans;
    if (anssofar == -1) anssofar = z;
    anssofar = min(anssofar, z);
  }
  anssofar = min(ans, anssofar);
  for (int i = 0; i < n; i++) {
    if (l > num[i]) continue;
    anssofar = min(cost[i], anssofar);
  }
  cout << anssofar;
  return 0;
}
