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.
题意：给你n 个数 按照升序排列, a1,a2,..an..分别代表2^a1...若个数加起来的结果等于( 2^u(u任意)-1 ) 则无需再添加数, 否则应计算出 需添加几个数
题解：假设最后加起来结果为2^k - 1,那么组成结果最少数量的数位，2^0,2^1,2^2,2^3,2^4...2^(k-1)，所以如果给出的数有相同的，他们也是无法共存的，因此我们可以把两个相同的数变为一个比他们大1的数，不断进行操作，优先队列就可以了，每次pop出两个数，看是否相等，相等就把 他们+1 的数 放进去，不想等就说明小的已经符合条件了，大的在放进去
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int n;
int main()
{
while(~scanf("%d",&n))
{
ll x;
priority_queue<ll,vector<ll>,greater<ll> >q;
for(int i=1;i<=n;i++)
{
scanf("%lld",&x);
q.push(x);
}
int ans=0;
while(q.size()>1)
{
ll m1=q.top();q.pop();
ll m2=q.top();q.pop();
if(m1==m2)
{
q.push(m1+1);
}
else ans++,q.push(m2);
}
x=q.top();
cout<<x-ans<<endl;
}
return 0;
}