Alyona and Mex

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121333#problem/B

Description

Someone gave Alyona an array containing n positive integers a1, a2, ..., an. 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, ..., bn such that 1 ≤ bi ≤ ai for every 1 ≤ 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, ..., an (1 ≤ ai ≤ 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：

可以任意减小数组元素的大小，使得最小不出现的自然数最大;

题解：

排序后离散化即可；当出现不能离散化时即为最大的自然数.

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>

#include <map>

#include <set>

#include <vector>

#define LL long long

#define eps 1e-8

#define maxn 110000

#define inf 0x3f3f3f3f

#define IN freopen("in.txt","r",stdin);

using namespace std;

int n;

int a[maxn];

int main(int argc, char const *argv[])

{

    //IN;

    while(scanf("%d",&n) != EOF)

    {

        for(int i=1; i<=n; i++)

            scanf("%d", &a[i]);

        sort(a+1,a+1+n);

        int ans = 0;

        for(int i=1; i<=n; i++)

            if(a[i] > ans) ans++;

        printf("%d\n", ans+1);

    }

    return 0;

}