#include <bits/stdc++.h>
using namespace std;
struct par {
  int v, k;
  par() {}
  par(const int& _v, const int& _k) : v(_v), k(_k) {}
};
int niz[100005];
int n;
deque<par> dq;
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &niz[i]);
  sort(niz, niz + n);
  for (int i = 0; i < n; i++) {
    int j = i;
    int v = niz[i], k = 0;
    for (; niz[j] == niz[i] && j < n; j++) k++;
    dq.push_back(par(v, k));
    i = j - 1;
  }
  vector<int> V;
  while (true) {
    if (dq.empty()) break;
    int v = dq[0].v, k = dq[0].k;
    dq.pop_front();
    int visak = k / 2;
    if (k % 2) V.push_back(v);
    if (!dq.empty() && dq[0].v == v + 1)
      dq[0].k += visak;
    else if (visak)
      dq.push_front(par(v + 1, visak));
  }
  int sol = V.back() - V.size() + 1;
  printf("%d\n", sol);
  return 0;
}

### Prompt

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

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;
struct par {
  int v, k;
  par() {}
  par(const int& _v, const int& _k) : v(_v), k(_k) {}
};
int niz[100005];
int n;
deque<par> dq;
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &niz[i]);
  sort(niz, niz + n);
  for (int i = 0; i < n; i++) {
    int j = i;
    int v = niz[i], k = 0;
    for (; niz[j] == niz[i] && j < n; j++) k++;
    dq.push_back(par(v, k));
    i = j - 1;
  }
  vector<int> V;
  while (true) {
    if (dq.empty()) break;
    int v = dq[0].v, k = dq[0].k;
    dq.pop_front();
    int visak = k / 2;
    if (k % 2) V.push_back(v);
    if (!dq.empty() && dq[0].v == v + 1)
      dq[0].k += visak;
    else if (visak)
      dq.push_front(par(v + 1, visak));
  }
  int sol = V.back() - V.size() + 1;
  printf("%d\n", sol);
  return 0;
}

```