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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
const long long int inf = 9e18;
const long double pi = 2 * acos(0.0);
using namespace std;
long long int power(long long int a, long long int n) {
  if (n == 0) {
    return 1;
  }
  long long int p = power(a, n / 2) % 1000000007;
  p = (p * p) % 1000000007;
  if (n % 2 == 1) {
    p = (p * a) % 1000000007;
  }
  return p % 1000000007;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
const int N = 1e6 + 5;
vector<long long int> sieve(N, 0);
void si() {
  sieve[1] = 1;
  for (int i = 2; i < N; i++) {
    if (sieve[i] == 0) {
      sieve[i] = 1;
      for (int j = 2 * i; j < N; j += i) {
        sieve[j] = i;
      }
    }
  }
}
vector<long long int> dp(30, 0);
long long int n, L;
long long int solve(long long int val, long long int curr) {
  long long int bit = -1;
  for (long long int i = 29; i >= 0; i--) {
    if ((val) & (1 << i)) {
      bit = i;
      break;
    }
  }
  long long int curr1 = curr, curr2;
  if (bit != 29) {
    curr1 = curr + dp[bit + 1];
  } else {
    curr1 = 1e18;
  }
  curr += dp[bit];
  val = val - (1 << bit);
  if (val != 0) {
    curr2 = solve(val, curr);
  } else {
    curr2 = curr;
  }
  return min(curr1, curr2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> L;
  vector<long long int> a(n);
  ;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  ;
  dp[0] = a[0];
  for (int i = 1; i < n; i++) {
    dp[i] = min(a[i], 2 * dp[i - 1]);
  }
  for (int i = n; i < 30; i++) {
    dp[i] = 2 * dp[i - 1];
  }
  long long int mini = 1e18;
  for (int i = 29; i >= 0; i--) {
    if (mini < dp[i]) {
      dp[i] = mini;
    }
    mini = min(mini, dp[i]);
  }
  long long int ans = solve(L, 0);
  cout << ans << endl;
  return 0;
}
