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 len, sum;
long long num[35];
long long dfs(long long has, long long val, int ind, int nee) {
  if ((has + (1 << ind + 1)) < nee) return (long long)1e18;
  if (has >= nee) return val;
  if (ind == -1) {
    if (has >= nee) return val;
    return (long long)1e18;
  }
  return min(dfs(has + (1 << ind), val + num[ind], ind - 1, nee),
             dfs(has, val, ind - 1, nee));
}
int main() {
  scanf("%lld%lld", &len, &sum);
  for (int i = 0; i < len; i++) {
    scanf("%lld", &num[i]);
  }
  long long now = num[0];
  for (int i = 1; i < len; i++) {
    now <<= 1;
    if (now < num[i])
      num[i] = now;
    else
      now = num[i];
  }
  long long sum2 = sum;
  long long xxx = sum / (1 << (len - 1)) - 1;
  long long ans = 0;
  if (xxx > 0) {
    ans += xxx * num[len - 1];
    sum -= xxx * (1 << (len - 1));
  }
  ans += dfs(0, 0, len - 1, sum);
  long long xxx2 = sum2 / (1 << (len - 1));
  long long ans2 = 0;
  if (xxx2 > 0) {
    ans2 += xxx2 * num[len - 1];
    sum2 -= xxx2 * (1 << (len - 1));
  }
  ans2 += dfs(0, 0, len - 1, sum2);
  printf("%lld\n", min(ans, ans2));
  return 0;
}
