#include <bits/stdc++.h>
using namespace std;
vector<long long> A;
struct node {
  long long u;
  bool operator<(const node& rhs) const { return u > rhs.u; }
};
int main() {
  int n;
  while (~scanf("%d", &n)) {
    priority_queue<node> Q;
    long long aa;
    node tt;
    for (int i = 1; i <= n; i++) {
      scanf("%I64d", &aa);
      tt.u = aa;
      Q.push(tt);
    }
    A.clear();
    node a;
    a = Q.top();
    Q.pop();
    node b;
    while (!Q.empty()) {
      b = Q.top();
      Q.pop();
      if (a.u == b.u) {
        b.u++;
        Q.push(b);
        a = Q.top();
        Q.pop();
        continue;
      }
      A.push_back(a.u);
      a = b;
    }
    A.push_back(a.u);
    aa = -1;
    long long ans = 0;
    for (int i = 0; i < A.size(); i++) {
      ans += A[i] - aa - 1;
      aa = A[i];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
vector<long long> A;
struct node {
  long long u;
  bool operator<(const node& rhs) const { return u > rhs.u; }
};
int main() {
  int n;
  while (~scanf("%d", &n)) {
    priority_queue<node> Q;
    long long aa;
    node tt;
    for (int i = 1; i <= n; i++) {
      scanf("%I64d", &aa);
      tt.u = aa;
      Q.push(tt);
    }
    A.clear();
    node a;
    a = Q.top();
    Q.pop();
    node b;
    while (!Q.empty()) {
      b = Q.top();
      Q.pop();
      if (a.u == b.u) {
        b.u++;
        Q.push(b);
        a = Q.top();
        Q.pop();
        continue;
      }
      A.push_back(a.u);
      a = b;
    }
    A.push_back(a.u);
    aa = -1;
    long long ans = 0;
    for (int i = 0; i < A.size(); i++) {
      ans += A[i] - aa - 1;
      aa = A[i];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```