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;
int64_t mod = 1e17 + 7;
int64_t mod1 = 1e9 + 6;
int64_t power(int64_t a, int64_t b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power((((a % mod) * (a % mod)) % mod), b / 2) % mod;
  else
    return ((a % mod) * (power((((a % mod) * (a % mod)) % mod), b / 2) % mod)) %
           mod;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t x[40];
  int64_t n, p;
  cin >> n >> p;
  int64_t a[51 + 1];
  for (int64_t i = 0; i < n; ++i) cin >> a[i];
  for (int64_t i = n; i < 40; ++i) a[i] = 1e18;
  x[0] = a[0];
  for (int64_t i = 1; i < 40; ++i) {
    for (int64_t j = 0; j < i; ++j) {
      a[i] = min(a[i], power(2, (i - j)) * a[j]);
    }
  }
  int64_t ans = 0;
  vector<int64_t> v;
  while (p) {
    v.push_back(p % 2);
    p /= 2;
  }
  int64_t prev = 0;
  for (int64_t i = 0; i < v.size(); ++i) {
    if (v[i])
      ans += a[i];
    else {
      ans = min(ans, a[i]);
    }
  }
  for (int64_t i = v.size(); i < 31; ++i) ans = min(ans, a[i]);
  cout << ans;
}
