B - Alyona and Mex

Description

Someone gave Alyona an array containing n positive integers a₁, a₂, ..., a_n. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all.

Formally, after applying some operations Alyona will get an array of n positive integers b₁, b₂, ..., b_n such that 1 ≤ b_i ≤ a_i for every1 ≤ i ≤ n. Your task is to determine the maximum possible value of mex of this array.

Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of elements in the Alyona's array.

The second line of the input contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the elements of the array.

Output

Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations.

Sample Input

Input

5
1 3 3 3 6

Output

5

Input

2
2 1

Output

3

题意：

给出n个元素，元素可交换或减小，求最终缺少的最小正整数的最大值（mex）。

仔细观察我们不难看出当n个元素为从1~n递增时mex取最大值，如n=5时，1,2,3,4,5，mex取最大值6。所以我们就将数列尽量接近连续递增数列。

如1 3 3 3 6可改为1 2 3 3 4则取最大值mex为5。

附AC代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;

 const int MAX=;

 int a[MAX];

 int main(){
     int n;
     while(cin>>n){
         for(int i=;i<n;i++){
             cin>>a[i];
         }
         int ans=;
         sort(a,a+n);//排序
         for(int i=;i<n;i++){
             if(a[i]>=ans)//注意等于
             ans++;
         }
         cout<<ans<<endl;
     }
     return ;
 }