Ivan has an array consisting of n elements. Each of the elements is an integer from 1 to n.

Recently Ivan learned about permutations and their lexicographical order. Now he wants to change (replace) minimum number of elements in his array in such a way that his array becomes a permutation (i.e. each of the integers from 1 to n was encountered in his array exactly once). If there are multiple ways to do it he wants to find the lexicographically minimal permutation among them.

Thus minimizing the number of changes has the first priority, lexicographical minimizing has the second priority.

In order to determine which of the two permutations is lexicographically smaller, we compare their first elements. If they are equal — compare the second, and so on. If we have two permutations x and y, then x is lexicographically smaller if x_i < y_i, where i is the first index in which the permutations x and y differ.

Determine the array Ivan will obtain after performing all the changes.

The first line contains an single integer n (2 ≤ n ≤ 200 000) — the number of elements in Ivan's array.

The second line contains a sequence of integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n) — the description of Ivan's array.

In the first line print q — the minimum number of elements that need to be changed in Ivan's array in order to make his array a permutation. In the second line, print the lexicographically minimal permutation which can be obtained from array with q changes.

In the first example Ivan needs to replace number three in position 1 with number one, and number two in position 3 with number four. Then he will get a permutation [1, 2, 4, 3] with only two changed numbers — this permutation is lexicographically minimal among all suitable.

In the second example Ivan does not need to change anything because his array already is a permutation.

/*

 * 题意：给你一个序列，元素范围[1,n],有重复的，问你最少更换几个数字，使得

 *  这个序列变成一个1~n的排列，并且字典序最小

 *

 *

 * 思路: 记录一下每个元素出现的个数，有多少没出现的，就要更换多少，然后从

 *  头开始替换，有限替换成字典序小的字符，如果要替换的字符，还要小，就加个

 *  标记，先不替换，最后再替换，这种策略保证了字典序最小

 *

 *

 * */

#include <bits/stdc++.h>

#define MAXN 200005

using namespace std;

int n;

int a[MAXN];

int vis[MAXN];//记录每个数字出现的次数

bool is[MAXN];

int cnt;

int pos;

int num;

inline void init(){

    memset(vis,,sizeof vis);

    memset(is,false,sizeof is);

    pos=;

    cnt=;

}

int main(){

    init();

    scanf("%d",&n);

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

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

        vis[a[i]]++;

    }

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

        if(vis[i]==)

            cnt++;

    printf("%d\n",cnt);

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

        if(vis[a[i]]>){

            while(vis[pos]!=)

                pos++;

            if(a[i]>pos){

                vis[a[i]]--;

                a[i]=pos;

                pos++;

            }else{

                if(is[a[i]]==true){

                    vis[a[i]]--;

                    a[i]=pos;

                    pos++;

                }else{

                    is[a[i]]=true;

                }

            }

        }

    }

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

        printf(i?" %d":"%d",a[i]);

    }

    puts("");

    return ;

}