#include <iostream>
#include <stdio.h>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <ctime>
#include <cassert>
#include <unordered_map>
#include <fstream>
#include <random>
#include <cstring>
#include <complex>
#include <bitset>
 
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define pb push_back
 
using namespace std;
 
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef double ld;

const int MAXN = 1e5 + 10;
const int INF = 1e9;

int dp[MAXN][35];

int main() {
    //ifstream cin("input.txt");
    //freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i)
        cin >> a[i];

    for (int j = 0; j < 32; ++j) {
        dp[0][j] = INF;
    }
    dp[0][0] = 0;

    for (int i = 0; i < n; ++i) {
        int odd = 0, even = 0;
        for (int j = 0; j < 32; ++j) {
            dp[i + 1][j] = INF;
        }

        for (; a[i] > 1; ) {
            if (a[i] & 1) {
                int pw = 1;
                for (; pw < a[i]; pw *= 2);
                pw /= 2;
                a[i] = pw;
                ++odd;
            } else {
                ++even;
                a[i] = a[i] / 2;
            }
        }

        for (int j = 0; j < 32; ++j) {
            if (odd == 0) {
                int cnt = j;
                int cur = dp[i][j];
                cur += max(0, even - cnt);
                cnt = even;
                dp[i + 1][cnt] = min(dp[i + 1][cnt], cur);
            } else {
                if (j > 0) {
                    int cnt = j - 1;
                    int cur = dp[i][j];
                    cur += max(0, even - cnt);
                    cnt = even;
                    dp[i + 1][cnt] = min(dp[i + 1][cnt], cur);
                }
                int cnt = j;
                int cur = dp[i][j] + 1;
                cur += max(0, even - cnt);
                cnt = even;
                ++cnt;
                dp[i + 1][cnt] = min(dp[i + 1][cnt], cur); 
            }
        }

        

        // int y = min(odd, cnt);
        // cnt -= y;
        // odd -= y;
        // if (cnt == 0) {
        //     cnt = even + odd;
        //     ans += cnt;
        // } else {
        //     if (even > cnt) {
        //         ans += even - cnt;
        //         even = cnt;
        //     } else {
        //         cnt = even;
        //     }
        // }
        // cout << "!" << ans << endl;
    }

    int ans = INF;
    for (int i = 0; i < 32; ++i) {
        ans = min(ans, dp[n][i]);
    }

    cout << ans << endl;


    // int ans = 0;
    // for (; ; ) {
    //     bool f = false;
    //     for (int i = 0; i < n; ++i) {
    //         f |= (a[i] != 1);
    //     }
    //     if (!f)
    //         break;
    //     int cnt = 0;
    //     bool start = false;

    //     for (int i = 0; i < n; ++i) {
    //         if (a[i] == 1) {
    //             start = false;
    //             continue;
    //         }
    //         if (!start) {
    //             start = true;
    //             ++ans;
    //             continue;
    //         }
    //         if (a[i] & 1)  {
    //             start = false;
    //         }
    //     }
    //     for (int i = 0; i < n; ++i) 
    //         a[i] = (a[i] + 1) / 2;
    // }
    // cout << ans << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <iostream>
#include <stdio.h>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <ctime>
#include <cassert>
#include <unordered_map>
#include <fstream>
#include <random>
#include <cstring>
#include <complex>
#include <bitset>
 
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define pb push_back
 
using namespace std;
 
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef double ld;

const int MAXN = 1e5 + 10;
const int INF = 1e9;

int dp[MAXN][35];

int main() {
    //ifstream cin("input.txt");
    //freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i)
        cin >> a[i];

    for (int j = 0; j < 32; ++j) {
        dp[0][j] = INF;
    }
    dp[0][0] = 0;

    for (int i = 0; i < n; ++i) {
        int odd = 0, even = 0;
        for (int j = 0; j < 32; ++j) {
            dp[i + 1][j] = INF;
        }

        for (; a[i] > 1; ) {
            if (a[i] & 1) {
                int pw = 1;
                for (; pw < a[i]; pw *= 2);
                pw /= 2;
                a[i] = pw;
                ++odd;
            } else {
                ++even;
                a[i] = a[i] / 2;
            }
        }

        for (int j = 0; j < 32; ++j) {
            if (odd == 0) {
                int cnt = j;
                int cur = dp[i][j];
                cur += max(0, even - cnt);
                cnt = even;
                dp[i + 1][cnt] = min(dp[i + 1][cnt], cur);
            } else {
                if (j > 0) {
                    int cnt = j - 1;
                    int cur = dp[i][j];
                    cur += max(0, even - cnt);
                    cnt = even;
                    dp[i + 1][cnt] = min(dp[i + 1][cnt], cur);
                }
                int cnt = j;
                int cur = dp[i][j] + 1;
                cur += max(0, even - cnt);
                cnt = even;
                ++cnt;
                dp[i + 1][cnt] = min(dp[i + 1][cnt], cur); 
            }
        }

        

        // int y = min(odd, cnt);
        // cnt -= y;
        // odd -= y;
        // if (cnt == 0) {
        //     cnt = even + odd;
        //     ans += cnt;
        // } else {
        //     if (even > cnt) {
        //         ans += even - cnt;
        //         even = cnt;
        //     } else {
        //         cnt = even;
        //     }
        // }
        // cout << "!" << ans << endl;
    }

    int ans = INF;
    for (int i = 0; i < 32; ++i) {
        ans = min(ans, dp[n][i]);
    }

    cout << ans << endl;


    // int ans = 0;
    // for (; ; ) {
    //     bool f = false;
    //     for (int i = 0; i < n; ++i) {
    //         f |= (a[i] != 1);
    //     }
    //     if (!f)
    //         break;
    //     int cnt = 0;
    //     bool start = false;

    //     for (int i = 0; i < n; ++i) {
    //         if (a[i] == 1) {
    //             start = false;
    //             continue;
    //         }
    //         if (!start) {
    //             start = true;
    //             ++ans;
    //             continue;
    //         }
    //         if (a[i] & 1)  {
    //             start = false;
    //         }
    //     }
    //     for (int i = 0; i < n; ++i) 
    //         a[i] = (a[i] + 1) / 2;
    // }
    // cout << ans << endl;
}

```