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;
long long dp[31];
long long final;
long long fun(long long n, long long divi, long long num) {
  long long val = 1, flag = 0, cnt = 0, i;
  if (n == 0) return 0;
  if (n == 1) {
    long long mini;
    mini = dp[0];
    for (i = 1; i < num; i++) mini = min(mini, dp[i]);
    return mini;
  }
  while (val <= n) {
    cnt++;
    val *= 2;
    if (val == n) {
      flag = 1;
      break;
    }
  }
  if (flag == 1) return dp[cnt];
  cnt = 0;
  val = 1;
  while (val <= divi) {
    cnt++;
    val *= 2;
  }
  val /= 2;
  long long ans, mod, val2;
  ans = n / divi;
  mod = n % divi;
  if (ans == 0) {
    val2 = dp[cnt - 1];
    return min(val2, fun(n, divi / 2, num));
  } else
    return min((ans + 1) * dp[cnt - 1],
               (ans * dp[cnt - 1] + fun(mod, divi / 2, num)));
}
int main() {
  long long n, i, L, c[31], val, tmp, j;
  cin >> n >> L;
  for (i = 0; i < n; i++) scanf("%lld", &c[i]);
  dp[0] = c[0];
  if (n == 1) {
    printf("%lld\n", L * c[0]);
    return 0;
  }
  for (i = 1; i < n; i++) {
    dp[i] = c[i];
    for (j = 0; j < i; j++)
      dp[i] = min(dp[i], (dp[j] * (long long)pow(2, i - j)));
    for (j = i + 1; j < n; j++) dp[i] = min(dp[i], c[j]);
  }
  tmp = n;
  val = L;
  L = L / pow(2, n - 1);
  final = 0;
  if (L == 0) {
    final = min(dp[n - 1], fun(val, (long long)pow(2, n - 2), n));
  } else {
    final = min((L + 1) * dp[n - 1],
                (L * dp[n - 1] + fun(val - L * (long long)pow(2, n - 1),
                                     (long long)pow(2, n - 2), n)));
  }
  L = val;
  long long val2, mod;
  val = L / pow(2, 0);
  mod = L % (long long)pow(2, 0);
  if (mod == 0)
    final = min(final, val * dp[0]);
  else
    final = min(final, (val + 1) * dp[0]);
  val = L / pow(2, 1);
  mod = L % 2;
  if (mod == 0)
    val2 = val * dp[1];
  else
    val2 = min((val + 1) * dp[1], val * dp[1] + fun(L - val * 2, 1, n));
  final = min(final, val2);
  for (i = 2; i < n; i++) {
    val = L / pow(2, i);
    mod = L % (long long)pow(2, i);
    if (mod == 0)
      val2 = val * dp[i];
    else
      val2 = min((val + 1) * dp[i],
                 val * dp[i] + fun(L - val * (long long)pow(2, i),
                                   (long long)pow(2, i - 1), n));
    final = min(final, val2);
  }
  for (i = 0; i < n; i++) {
    val = L / pow(2, i);
    mod = L % (long long)pow(2, i);
    if (mod != 0)
      final = min(final, (val + 1) * dp[i]);
    else
      final = min(final, val * dp[i]);
  }
  printf("%lld\n", final);
  return 0;
}
