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 target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using lld = long double;
using l64 = int64_t;
const int oo = 0x3f3f3f3f;
const ll MOD = 1000000007;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ll n, k;
  cin >> n >> k;
  vector<ll> ar(n);
  for (ll i = 0; i < n; i++) cin >> ar[i];
  for (ll i = 1; i < n; i++) ar[i] = min(ar[i], ar[i - 1] * 2);
  ll ans = 0;
  uint64_t l = 0;
  uint64_t h = 2e18;
  while (l + 1 < h) {
    uint64_t mid = (l + h) >> 1;
    uint64_t sum = 0;
    uint64_t x = mid;
    for (ll i = n - 1; i >= 0; i--) {
      sum += (x / ar[i]) * (1LL << i);
      if (sum >= k) break;
      x %= ar[i];
    }
    if (sum >= k)
      h = mid;
    else
      l = mid;
  }
  cout << h << '\n';
  return 0;
}
