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 MAXN = 100;
const long long INF = 9e18;
const int LOG = 30;
long long c[MAXN];
long long dp[MAXN];
void solve() {
  int n, L;
  cin >> n >> L;
  for (int i = 0; i < n; i++) cin >> c[i];
  for (int i = 0; i <= LOG; i++) dp[i] = INF;
  for (int i = 0; i <= LOG; i++) {
    if (i < n) {
      dp[i] = c[i];
    }
    long long cur = 1;
    for (int j = i - 1; j >= 0; j--) {
      cur <<= 1;
      if (cur > dp[i]) break;
      dp[i] = min(dp[i], dp[j] * cur);
    }
  }
  long long ans = 0;
  for (int i = 0; i <= LOG; i++) {
    if (L & (1 << i)) {
      ans += dp[i];
    }
    if (0) cout << "DP[ " << i << "] = " << dp[i] << endl;
  }
  for (int rep = 0; rep <= LOG; rep++) {
    if (L & (1 << rep)) {
      continue;
    }
    long long l2 = L | (1 << rep);
    for (int j = 0; j < rep; j++) {
      if (l2 & (1 << j)) {
        l2 ^= 1 << j;
      }
    }
    long long cur = 0;
    for (int j = 0; j <= LOG; j++) {
      if (l2 & (1 << j)) {
        cur += dp[j];
      }
    }
    ans = min(ans, cur);
  }
  cout << ans;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
