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 cost[maxn];
long long equatedCost[maxn];
long long cap[maxn];
pair<int, int> getMinInd(int n, int L) {
  int minInd = 0;
  long long minCost = equatedCost[0];
  int i;
  for (i = 0; i < n; ++i) {
    if (cap[i] > L) break;
    if (equatedCost[i] < minCost) {
      minInd = i;
      minCost = equatedCost[i];
    }
  }
  if (i == n) return {minInd, -1};
  long long cost2 = (L / cap[minInd]) * cost[minInd];
  int cost3 = *min_element(cost + i, cost + n);
  if (cost2 < cost3)
    return {minInd, cost3};
  else
    return {-1, cost3};
}
int main() {
  int n, L;
  cin >> n >> L;
  long long pow2 = 1;
  long long maxPow2 = (1 << (n - 1));
  for (int i = 0; i < n; ++i) {
    cin >> cost[i];
    cap[i] = pow2;
    equatedCost[i] = (maxPow2 / pow2) * cost[i];
    pow2 *= 2;
  }
  long long ans = 2e18;
  long long ca = 0;
  while (L != 0) {
    pair<int, int> p1 = getMinInd(n, L);
    int ind = p1.first;
    int otherCost = p1.second;
    if (ind == -1) {
      ca += otherCost;
      L = 0;
    } else {
      long long posAns = ca + otherCost;
      if (otherCost != -1) ans = min(ans, posAns);
      int bottles = L / cap[ind];
      ca += bottles * cost[ind];
      L -= bottles * cap[ind];
    }
  }
  ans = min(ans, ca);
  cout << ans << "\n";
  return 0;
}
