#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
typedef long long LL;
typedef pair<int, int> PII;

int n;
int a[100000], b[100000];

int main() {
    scanf("%d", &n);
    forn(i, n) {
        int x;
        scanf("%d", &x);
        while (x > 1) {
            if (x % 2 == 1) b[i] = 1;
            ++a[i];
            x = (x + 1) / 2;
        }
    }
    int ans = a[0];
    for (int i = 1; i < n; ++i) {
        ans += max(0, a[i] - a[i - 1]);
    }
    forn(i, n) if (b[i]) {
        b[i] = 0;
        if (i > 0 && a[i] <= a[i - 1]) {
            if (i < n - 1 && a[i] <= a[i + 1]) {
                b[i] = 1;
            }
        }
    }
    forn(i, n - 1) if (b[i] == 1 && b[i + 1] == 1) {
        b[i] = 0;
        b[i + 1] = 0;
        ans += 1;
    }
    forn(i, n) ans += b[i];
    cout << ans << endl;
    return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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
#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
typedef long long LL;
typedef pair<int, int> PII;

int n;
int a[100000], b[100000];

int main() {
    scanf("%d", &n);
    forn(i, n) {
        int x;
        scanf("%d", &x);
        while (x > 1) {
            if (x % 2 == 1) b[i] = 1;
            ++a[i];
            x = (x + 1) / 2;
        }
    }
    int ans = a[0];
    for (int i = 1; i < n; ++i) {
        ans += max(0, a[i] - a[i - 1]);
    }
    forn(i, n) if (b[i]) {
        b[i] = 0;
        if (i > 0 && a[i] <= a[i - 1]) {
            if (i < n - 1 && a[i] <= a[i + 1]) {
                b[i] = 1;
            }
        }
    }
    forn(i, n - 1) if (b[i] == 1 && b[i + 1] == 1) {
        b[i] = 0;
        b[i + 1] = 0;
        ans += 1;
    }
    forn(i, n) ans += b[i];
    cout << ans << endl;
    return 0;
}

```