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;
inline void ReadInt(int &num) {
  num = 0;
  int f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    num = num * 10 + (c - '0');
    c = getchar();
  }
  num = num * f;
  return;
}
inline void ReadLong(long long &num) {
  num = 0;
  int f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    num = num * 10 + (c - '0');
    c = getchar();
  }
  num = num * f;
  return;
}
int n, l;
int v[32], c[32], p[32];
long long dfs(int cur) {
  if (!cur) return 0;
  int i, j;
  long long mn = 1e18;
  for (i = 0; i < n && v[i] <= cur; i++)
    ;
  i--;
  for (j = i + 1; j < n; j++) mn = min(mn, (long long)c[j]);
  return min(mn, (long long)cur / v[p[i]] * c[p[i]] + dfs(cur % v[p[i]]));
}
int main() {
  int cnt = 1;
  cin >> n >> l;
  for (int i = 0; i < n; i++) {
    cnt *= 2;
    v[i] = cnt / 2;
    cin >> c[i];
    if (!i)
      p[i] = i;
    else if (c[p[i - 1]] * (v[i - p[i - 1]]) < c[i])
      p[i] = p[i - 1];
    else
      p[i] = i;
  }
  cout << dfs(l) << endl;
  return 0;
}
