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 xi < yi, 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, ..., an (1 ≤ ai ≤ 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.

算法：思维

题意：给你一个长度为n的数组，里面有重复的元素，你需要把这个多余的重复元素改成1 ~ n中那些你没有用过的元素，问你需要更改多少次，以及最小的元素序列是什么？

思路：首先，我先将那些多余的重复元素和那些没有用过的元素记录下来，然后就进行遍历判断。假如当前元素时重复元素，并且当前元素比未使用过的最小元素大的话，就将其覆盖，否则，你就需要记录当前元素已经使用，当之后在遇到这个元素的时候，我就可以直接覆盖。

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <set>

#include <map>

using namespace std;

#define INF 0x3f3f3f3f

typedef long long ll;

const int maxn = 2e5+;

int vis[maxn];

int v[maxn];

int n;

int arr[maxn];

int b[maxn];

int main() {

    scanf("%d", &n);

    int k = ;

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

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

        vis[arr[i]]++;      //记录使用过的每个元素的个数

    }

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

        if(!vis[i]) {

            b[k++] = i;     //记录那些没有使用的元素

        }

    }

    int j = ;

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

        int x = arr[i];

        if(vis[x] > ) {        //当使用过的元素有多个时

            if(b[j] < x) {      //如果当前未使用的最小元素比其小，那么直接覆盖

                arr[i] = b[j++];

                vis[x]--;

            } else {

                if(v[x]) {      //当前元素已被记录，在它前面有个和它一样的

                    arr[i] = b[j++];

                }

                v[x]++;

            }

        }

    }

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

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

        printf("%d%c", arr[i], " \n"[i == n]);

    }

    return ;

}