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 long long int INF = 1600000000000000000LL;
const int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1};
const int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int dx4[] = {0, 0, 1, -1};
const int dy4[] = {1, -1, 0, 0};
const int MOD = 1e9 + 7;
const int MODB = 1e9 + 9;
const int N = 35;
int invalid[N];
int c[N];
int n;
long long int L;
vector<pair<long long int, long long int> > v;
int main() {
  cin >> n >> L;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  for (int i = 1; i <= n; i++) {
    if (invalid[i]) continue;
    for (int j = i + 1; j <= n; j++) {
      if (c[j] <= c[i]) invalid[i] = 1;
      if (c[i] * 1LL * (1LL << (j - i)) <= c[j]) {
        invalid[j] = 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!invalid[i]) {
      v.push_back(make_pair(i - 1, c[i]));
    }
  }
  long long int s = 0;
  long long int ans = INF;
  int cx = 0;
  while (L > 0 && cx <= 10000) {
    int sz = v.size();
    for (int i = sz - 1; i >= 0; --i) {
      long long int cur = L / (1LL << v[i].first);
      if (cur == 0) {
        ans = min(ans, s + v[i].second);
      } else {
        s = s + cur * v[i].second;
        L -= (cur * 1LL * (1LL << v[i].first));
        break;
      }
    }
    cx++;
  }
  if (L <= 0) {
    ans = min(ans, s);
  }
  cout << ans;
}
