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 oo = 3e18;
const int MOD = 1000000007;
const int N = 500010;
int n, L, ATM;
long long a[N];
pair<long long, long long> dp[N];
long long buy(long long L) {
  long long cur = 0;
  for (long long i = 32; i >= 0; --i) {
    if (L >= (1ll << i)) {
      cur += dp[i].first;
      L -= (1ll << i);
    }
  }
  return cur;
}
long long ce(long long a, long long b) { return (a / b) + (a % b != 0); }
int main() {
  cin >> n >> L;
  ATM = L;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i <= 32; ++i) {
    dp[i] = make_pair(oo, (1ll << i));
  }
  dp[0].first = a[0];
  for (int i = 1; i < n; ++i) {
    dp[i] = min(make_pair(a[i], 1ll << i),
                make_pair(dp[i - 1].first * 2, 1ll << i));
  }
  for (int i = 1; i <= 32; ++i) {
    dp[i] = min(dp[i], make_pair(dp[i - 1].first * 2, 1ll << i));
  }
  for (int i = 31; i >= 0; --i) {
    if (dp[i + 1].first <= dp[i].first) dp[i] = dp[i + 1];
  }
  long long cur = 0;
  long long best = buy(L);
  for (long long i = 32; i >= 0; --i) {
    best = min(best, ce(ATM, dp[i].second) * dp[i].first);
  }
  for (long long i = 32; i >= 0; --i) {
    if ((1ll << i) >= L && buy(L) >= dp[i].first) {
      best = min(best, cur + dp[i].first);
    }
    if (L >= (1ll << i)) {
      cur += dp[i].first;
      L -= dp[i].second;
    }
  }
  cout << min(cur, best) << endl;
  return 0;
}
