#include <bits/stdc++.h>
using namespace std;
const int M = 1e3 + 228;
const long double Q = 4 * 1e9 + 7.;
const long double eps = 1e-10;
vector<pair<long long, int> > g;
int a[100 * M];
int main() {
  srand(time(NULL));
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int pos = 0;
  while (pos < n) {
    int i = pos;
    while (i < n && a[pos] == a[i]) i++;
    g.push_back(make_pair(a[pos], i - pos));
    pos = i;
  }
  reverse(g.begin(), g.end());
  long long cnt = 0;
  while (1) {
    if ((int)g.size() == 1 && g[0].second == 1) break;
    pair<long long, int> cur = g.back();
    g.pop_back();
    cnt += cur.second % 2;
    cur.second /= 2;
    if (cur.second != 0) {
      if ((int)g.size() && g.back().first == cur.first + 1)
        g.back().second += cur.second;
      else
        g.push_back(make_pair(cur.first + 1, cur.second));
    }
  }
  printf("%lld", g[0].first - cnt);
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

Ivan has got an array of n non-negative integers a1, a2, ..., an. Ivan knows that the array is sorted in the non-decreasing order. 

Ivan wrote out integers 2a1, 2a2, ..., 2an on a piece of paper. Now he wonders, what minimum number of integers of form 2b (b ≥ 0) need to be added to the piece of paper so that the sum of all integers written on the paper equalled 2v - 1 for some integer v (v ≥ 0). 

Help Ivan, find the required quantity of numbers.

Input

The first line contains integer n (1 ≤ n ≤ 105). The second input line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 2·109). It is guaranteed that a1 ≤ a2 ≤ ... ≤ an.

Output

Print a single integer — the answer to the problem.

Examples

Input

4
0 1 1 1


Output

0


Input

1
3


Output

3

Note

In the first sample you do not need to add anything, the sum of numbers already equals 23 - 1 = 7.

In the second sample you need to add numbers 20, 21, 22.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int M = 1e3 + 228;
const long double Q = 4 * 1e9 + 7.;
const long double eps = 1e-10;
vector<pair<long long, int> > g;
int a[100 * M];
int main() {
  srand(time(NULL));
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int pos = 0;
  while (pos < n) {
    int i = pos;
    while (i < n && a[pos] == a[i]) i++;
    g.push_back(make_pair(a[pos], i - pos));
    pos = i;
  }
  reverse(g.begin(), g.end());
  long long cnt = 0;
  while (1) {
    if ((int)g.size() == 1 && g[0].second == 1) break;
    pair<long long, int> cur = g.back();
    g.pop_back();
    cnt += cur.second % 2;
    cur.second /= 2;
    if (cur.second != 0) {
      if ((int)g.size() && g.back().first == cur.first + 1)
        g.back().second += cur.second;
      else
        g.push_back(make_pair(cur.first + 1, cur.second));
    }
  }
  printf("%lld", g[0].first - cnt);
  return 0;
}

```