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;
  long long c[35];
  scanf("%lld %lld", &n, &l);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
  }
  long long tot = 0, ans = -1;
  while (l) {
    long long t = 1;
    while ((t << 1) <= l) {
      t <<= 1;
    }
    long long mn = -1;
    long long i;
    for (i = 1; i <= n; i++) {
      if (t < (1 << (i - 1))) {
        break;
      }
      long long p = t / (1 << (i - 1));
      if (mn == -1) {
        mn = p * c[i];
      } else {
        mn = min(mn, p * c[i]);
      }
    }
    bool flag = false;
    for (; i <= n; i++) {
      if (ans == -1) {
        ans = c[i] + tot;
      } else {
        ans = min(ans, c[i] + tot);
      }
    }
    l -= t;
    tot += mn;
  }
  if (ans == -1) {
    ans = tot;
  } else {
    ans = min(ans, tot);
  }
  printf("%lld\n", ans);
}
