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;
const unsigned long long inf = 1e18;
unsigned long long n, l, c[105], x, ans, s, t, dp1[(1 << 16)], dp2[(1 << 16)],
    s1, s2;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l;
  s = 1e18;
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
    if (i > 0) c[i] = min(c[i], c[i - 1] * 2);
  }
  for (int i = n; i <= 31; ++i) {
    c[i] = c[i - 1] * 2;
  }
  for (int i = n; i >= 0; --i) {
    c[i] = min(c[i + 1], c[i]);
  }
  ans = c[31];
  for (long long i = 0; i < (1 << 16); ++i) {
    s1 = 0;
    s2 = 0;
    for (long long j = 0; j < 16; ++j) {
      if (i & (1 << j)) {
        s1 += c[j];
        s2 += c[j + 16];
        if (s1 > ans) s1 = ans;
        if (s2 > ans) s2 = ans;
      }
    }
    dp1[i] = s1;
    dp2[i] = s2;
  }
  for (int i = (1 << 16) - 2; i >= 0; --i) {
    dp1[i] = min(dp1[i], dp1[i + 1]);
    dp2[i] = min(dp2[i], dp2[i + 1]);
  }
  for (long long i = 0; i < (1 << 16); ++i) {
    x = i;
    x = (x << 16);
    if (x >= l)
      s1 = 0;
    else
      s1 = l - x;
    if (s1 < (1 << 16)) ans = min(ans, dp1[s1] + dp2[i]);
  }
  cout << ans;
}
