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 int MAXN = 32;
long long N, L;
long long cost[MAXN];
void input() {
  cin >> N >> L;
  cin >> cost[0];
  for (int i = 1; i < N; i++) {
    cin >> cost[i];
    cost[i] = min(cost[i], cost[i - 1] + cost[i - 1]);
  }
  for (int i = N - 2; i >= 0; i--) cost[i] = min(cost[i], cost[i + 1]);
}
long long solve(long long x) {
  if (x == 0) return 0;
  long long maxPow = 0;
  for (int i = 0; i < N; i++)
    if ((1 << i) <= x) maxPow = i;
  long long twoPow = 1 << maxPow;
  long long ret = solve(x % twoPow) + (cost[maxPow] * (x / twoPow));
  if (maxPow != N - 1) return min(ret, cost[maxPow + 1]);
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  input();
  cout << solve(L);
  return 0;
}
