#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 100005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N;
int a[MAXN],cnt[MAXN],val[MAXN][2];
int64 dp[MAXN][2];
void Solve() {
	read(N);
	for(int i = 1 ; i <= N ; ++i) {
		read(a[i]);
		int k = -1,x = a[i];
		while(x) {++k;x >>= 1;}
		cnt[i] = k;
	}
	for(int i = 1 ; i <= N ; ++i) {
		dp[i][0] = 1e16;dp[i][1] = 1e16;
		int s = cnt[i],t = cnt[i];
		if(a[i] != (1 << cnt[i])) ++t;
		val[i][0] = s;val[i][1] = t;
		for(int j = 0 ; j <= 1 ; ++j) {
			dp[i][0] = min(dp[i][0],dp[i - 1][j] + t - min(t,val[i - 1][j]));
			dp[i][1] = min(dp[i][1],dp[i - 1][j] + t - min(s,val[i - 1][j]));
		}
	}
	out(min(dp[N][0],dp[N][1]));enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

### Prompt

Create a solution in cpp for 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
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 100005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N;
int a[MAXN],cnt[MAXN],val[MAXN][2];
int64 dp[MAXN][2];
void Solve() {
	read(N);
	for(int i = 1 ; i <= N ; ++i) {
		read(a[i]);
		int k = -1,x = a[i];
		while(x) {++k;x >>= 1;}
		cnt[i] = k;
	}
	for(int i = 1 ; i <= N ; ++i) {
		dp[i][0] = 1e16;dp[i][1] = 1e16;
		int s = cnt[i],t = cnt[i];
		if(a[i] != (1 << cnt[i])) ++t;
		val[i][0] = s;val[i][1] = t;
		for(int j = 0 ; j <= 1 ; ++j) {
			dp[i][0] = min(dp[i][0],dp[i - 1][j] + t - min(t,val[i - 1][j]));
			dp[i][1] = min(dp[i][1],dp[i - 1][j] + t - min(s,val[i - 1][j]));
		}
	}
	out(min(dp[N][0],dp[N][1]));enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

```