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;
double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005LL;
double PI = acos(-1);
int dirx[2] = {0, 1};
int diry[2] = {1, 0};
long long int MOD = 1000000007;
long long int dfs(int node, int par, vector<vector<int> > &adj,
                  vector<long long int> &counter, int n) {
  long long int count = 1, kl;
  for (int i = 0; i < adj[node].size(); i++) {
    if (adj[node][i] != par) {
      count += dfs(adj[node][i], node, adj, counter, n);
    }
  }
  kl = ((long long int)n - count);
  counter[node] = (count * kl);
  return count;
}
bool rule(pair<long long int, long long int> a,
          pair<long long int, long long int> b) {
  return a.first < b.first;
}
long long int min(long long int a, long long int b) {
  if (a > b) {
    return b;
  }
  return a;
}
int main() {
  int t;
  t = 1;
  while (t--) {
    int n, x;
    cin >> n >> x;
    vector<long long int> arr(n);
    long long int mod = 1000000001;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    for (int i = 0; i < n; i++) {
      long long int y = arr[i];
      for (int j = i + 1; j < n; j++) {
        y *= 2;
        if (y > mod) {
          break;
        }
        arr[j] = min(arr[j], y);
      }
    }
    for (int i = n; i <= 30; i++) {
      arr.push_back((arr[i - 1] * 2));
    }
    int pos = 0;
    unordered_set<int> op;
    int count = 0;
    long long int maxi = 0, p = x;
    while (p) {
      if (p % 2) {
        maxi += arr[pos];
        count++;
        op.insert(pos);
      }
      pos++;
      p /= 2;
    }
    pos = 30;
    while (op.find(pos) == op.end()) {
      maxi = min(maxi, arr[pos]);
      pos--;
    }
    long long int ui = 0;
    while (pos >= 0) {
      long long int h = ui;
      if (op.find(pos) != op.end()) {
        ui += arr[pos];
        h += arr[pos];
      }
      count--;
      if (count == 0) {
        maxi = min(maxi, ui);
        break;
      }
      maxi = min(maxi, (ui + arr[pos]));
      pos--;
      while (op.find(pos) == op.end()) {
        maxi = min(maxi, (ui + arr[pos]));
        pos--;
      }
    }
    cout << maxi << endl;
  }
}
