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 comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int MOD = 10;
const long double E = 1e-9;
unsigned char ccc;
bool _minus = false;
template <typename T>
inline T sqr(T t) {
  return (t * t);
}
inline void read(long long &n) {
  n = 0;
  _minus = false;
  while (true) {
    ccc = getchar();
    if (ccc == ' ' || ccc == '\n') break;
    if (ccc == '-') {
      _minus = true;
      continue;
    }
    n = n * 10 + ccc - '0';
  }
  if (_minus) n *= -1;
}
inline bool read(int &n) {
  n = 0;
  _minus = false;
  while (true) {
    ccc = getchar();
    if (ccc == ' ' || ccc == '\n') {
      if (ccc == '\n') return true;
      break;
    }
    if (ccc == '-') {
      _minus = true;
      continue;
    }
    n = n * 10 + ccc - '0';
  }
  if (_minus) n *= -1;
  return false;
}
char wwww[19];
int kkkk;
inline void write(long long y) {
  long long x = y;
  kkkk = 0;
  if (x < 0) {
    putchar('-');
    x *= -1;
  }
  if (!x) {
    ++kkkk;
    wwww[kkkk] = '0';
  } else
    while (x) {
      ++kkkk;
      wwww[kkkk] = char(x % 10 + '0');
      x /= 10;
    }
  for (int i = kkkk; i >= 1; --i) putchar(wwww[i]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  srand((unsigned int)time(NULL));
  cout.precision(10);
  cout << fixed;
  int n, l;
  cin >> n >> l;
  long long ans = 1e18;
  vector<long long> dp(n);
  for (int i = 0; i < n; i++) {
    cin >> dp[i];
  }
  for (int i = 1; i < n; i++) {
    dp[i] = min(dp[i], dp[i - 1] * 2);
  }
  for (int i = n - 2; i >= 0; i--) {
    dp[i] = min(dp[i], dp[i + 1]);
  }
  int max_val = (1 << (n - 1));
  long long add = (l / max_val) * 1LL * dp[n - 1];
  l %= max_val;
  if (l == 0) {
    ans = 0;
  } else {
    ans = dp[n - 1];
    long long e = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (l & (1 << i)) {
        e += dp[i];
      }
      ans = min(ans, e + dp[i]);
    }
    ans = min(ans, e);
  }
  cout << (ans + add) << "\n";
}
