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 int mod = 1e9 + 7;
const double error = 1e-10;
const double PI = acos(-1);
inline long long int MOD(long long int x, long long int m = mod) {
  long long int y = x % m;
  return (y >= 0) ? y : y + m;
}
const int inf = INT_MAX;
const long long int infl = LLONG_MAX;
const int nmax = 2e3 + 10;
long long int cost[35];
int main() {
  int n;
  long long int v;
  scanf("%d %lld", &n, &v);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &cost[i]);
  }
  for (int i = 1; i <= 30; i++) {
    long long int x = cost[i - 1] << 1;
    if (cost[i] == 0)
      cost[i] = x;
    else
      cost[i] = min(cost[i], x);
  }
  long long int mn = infl;
  for (int i = 30; i >= 0; i--) {
    long long int temp = 0;
    for (int j = 30; j > i; j--) {
      if (v & (1 << j)) temp += cost[j];
    }
    if (!(v & (1 << i))) {
      temp += cost[i];
      mn = min(mn, temp);
    }
  }
  long long int temp = 0;
  for (int i = 0; i <= 30; i++)
    if (v & (1 << i)) temp += cost[i];
  mn = min(mn, temp);
  printf("%lld", mn);
}
