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 N = 50;
using ll = long long;
using pii = pair<ll, int>;
const ll inf = 0x3f3f3f3f3f3f3f3fLL;
int n, gap;
ll L, c[N], ans, hop;
bool bit[N];
vector<int> p;
bool f;
int main() {
  cin >> n >> L;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", c + i);
    if (!f) {
      hop = c[i];
      gap = i;
      f = 1;
      continue;
    }
    ll tmp = (ll)(1 << (i - gap)) * hop;
    if (tmp < c[i]) {
      c[i] = tmp;
    } else {
      hop = c[i];
      gap = i;
    }
  }
  for (int i = n; i <= 30; ++i) {
    c[i] = (ll)(1 << (i - gap)) * hop;
  }
  int cnt = 0;
  while (L) {
    if (L % 2 == 1) {
      p.push_back(cnt);
      ans += c[cnt];
    }
    bit[cnt++] = L % 2;
    L >>= 1;
  }
  reverse(p.begin(), p.end());
  for (int i = p[0] + 1; i <= 30; ++i) {
    ans = min(ans, c[i]);
  }
  ll base = c[p[0]];
  for (int i = 1; i < p.size(); ++i) {
    for (int j = p[i] + 1; j < p[i - 1]; ++j) {
      ans = min(ans, base + c[j]);
    }
    base += c[p[i]];
  }
  cout << ans << endl;
}
