// ※※※ 解答不能 ※※※
// ztzshiwo氏.
// https://atcoder.jp/contests/code-festival-2017-quala/submissions/1779879
#include <bits/stdc++.h>
using namespace std;
#define B2(x) (63 - __builtin_clzll(x))
const int N = 100007;
int n, ans, l[N], r[N], dp[N][2];
int main(){
    int x;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        l[i] = B2(x);
        r[i] = l[i] + ((1 << l[i]) != x);
        ans += r[i];
        dp[i][0] = max(dp[i - 1][0] + min(r[i - 1], l[i]), dp[i - 1][1] + min(l[i - 1], l[i]));
        dp[i][1] = max(dp[i - 1][0] + min(r[i - 1], r[i]), dp[i - 1][1] + min(l[i - 1], r[i]));
    }
    printf("%d\n", ans - max(dp[n][0], dp[n][1]));
    return 0;
}
### Prompt

In Cpp, your task is to solve the following problem:

There are A slimes lining up in a row. Initially, the sizes of the slimes are all 1.

Snuke can repeatedly perform the following operation.

* Choose a positive even number M. Then, select M consecutive slimes and form M / 2 pairs from those slimes as follows: pair the 1-st and 2-nd of them from the left, the 3-rd and 4-th of them, ..., the (M-1)-th and M-th of them. Combine each pair of slimes into one larger slime. Here, the size of a combined slime is the sum of the individual slimes before combination. The order of the M / 2 combined slimes remain the same as the M / 2 pairs of slimes before combination.



Snuke wants to get to the situation where there are exactly N slimes, and the size of the i-th (1 ≤ i ≤ N) slime from the left is a_i. Find the minimum number of operations required to achieve his goal.

Note that A is not directly given as input. Assume A = a_1 + a_2 + ... + a_N.

Constraints

* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the minimum number of operations required to achieve Snuke's goal.

Examples

Input

2
3 3


Output

2


Input

4
2 1 2 2


Output

2


Input

1
1


Output

0


Input

10
3 1 4 1 5 9 2 6 5 3


Output

10

### Response

```cpp
// ※※※ 解答不能 ※※※
// ztzshiwo氏.
// https://atcoder.jp/contests/code-festival-2017-quala/submissions/1779879
#include <bits/stdc++.h>
using namespace std;
#define B2(x) (63 - __builtin_clzll(x))
const int N = 100007;
int n, ans, l[N], r[N], dp[N][2];
int main(){
    int x;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        l[i] = B2(x);
        r[i] = l[i] + ((1 << l[i]) != x);
        ans += r[i];
        dp[i][0] = max(dp[i - 1][0] + min(r[i - 1], l[i]), dp[i - 1][1] + min(l[i - 1], l[i]));
        dp[i][1] = max(dp[i - 1][0] + min(r[i - 1], r[i]), dp[i - 1][1] + min(l[i - 1], r[i]));
    }
    printf("%d\n", ans - max(dp[n][0], dp[n][1]));
    return 0;
}
```