Problem:
Score : 1600 points

Problem Statement

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.

Sample Input 1

2
3 3

Sample Output 1

2

One way to achieve Snuke's goal is as follows.
Here, the selected slimes are marked in bold.

(1, 1, 1, 1, 1, 1) → (1, 2, 2, 1)

(1, 2, 2, 1) → (3, 3)

Sample Input 2

4
2 1 2 2

Sample Output 2

2

One way to achieve Snuke's goal is as follows.

(1, 1, 1, 1, 1, 1, 1) → (2, 1, 1, 1, 1, 1)

(2, 1, 1, 1, 1, 1) → (2, 1, 2, 2)

Sample Input 3

1
1

Sample Output 3

0

Sample Input 4

10
3 1 4 1 5 9 2 6 5 3

Sample Output 4

10

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <map>
using namespace std;

int n, a[110000], ans;
int b[110000];
int dp[110000][33];

int get(int x) {
	if (x == 1)
		return 0;
	int l = -1, r = 31;
	while (l < r - 1) {
		int mid = (l + r) / 2;
		if ((1 << mid) >= x)
			r = mid;
		else
			l = mid;
	}
	return r;
}

void up(int &x, int y) {
	x = min(x, y);
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		b[i] = get(a[i]);
	}
	dp[0][0] = 0;
	for (int i = 1; i <= 32; i++)
		dp[0][i] = 1e9;
	for (int i = 1; i <= n; i++)
		for (int j = 0; j <= 32; j++)
			dp[i][j] = 1e9;
	for (int i = 1; i <= n; i++)
		for (int j = 0; j <= 32; j++) {
			int p = j, cost = dp[i - 1][j];
			if (a[i] % 2 == 1) {
				if (p > 0) {
					p -= 1;
					up(dp[i][max(0, b[i] - 1)], cost + max(0, b[i] - 1 - p));
					p += 1;
				}
				up(dp[i][b[i]], cost + max(1, max(0, b[i] - p)));
			}else {
				up(dp[i][b[i]], cost + max(0, b[i] - p));
			}
		}
	ans = 1e9;
	for (int i = 0; i <= 32; i++)
		ans = min(ans, dp[n][i]);
	printf("%d\n", ans);
}