codeforce864d

D. Make a Permutation!

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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.

Output

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.

Examples

input

4
3 2 2 3

output

2
1 2 4 3 

input

6
4 5 6 3 2 1

output

0
4 5 6 3 2 1 

input

10
6 8 4 6 7 1 6 3 4 5

output

3
2 8 4 6 7 1 9 3 10 5 

Note

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.

输入n    输入n个数字    这n个数字可能会有重复的   让你进行替换   结果要变成由1-n排列的n个数字  

思路：贪心    记录一下每个数字出现的地方，我选择queue<int>  arr[2e5+5]

queue<int,vector<int>,greater<int> >  pp;  装没有出现过的数字

首先输入n个数字for(i = 1; i <= n; ++i)            cin >> ans[i];

for(i = 1; i <= n; ++i) 当arr[ans[i]].size() == 1那就跳过    如果大于1

否则就判断一下  pp.top()是否大于当前的ans[i]，小于就替换

大于等于就不替换   下次再扫描到相同的数值时再替换

#include <iostream>

#include <queue>

#include <string>

#include <cstring>

using namespace std;

const int maxn = 2e5+5;

queue<int> ans[maxn];

int main()

{

    int n,i,j,m,sum = 0;

    priority_queue<int,vector<int>,greater<int> > pp;

    int arr[maxn];

    bool pan[maxn];

    memset(pan, 0, sizeof(pan));

    

    

    scanf("%d",&n);

    for(i = 1; i <= n; ++i)

    {

        scanf("%d",arr+i);

        ans[arr[i]].push(i);

    }

    

    for(i = 1; i <= n; ++i)

    {

        if(ans[i].size() == 0)

            pp.push(i);

        if(ans[i].size() > 1)

            sum += ans[i].size() - 1;

    }

    

    for(i = 1; i <= n; ++i)

    {

        if(ans[arr[i]].size() == 1)

            continue;

        

        if(pp.top() < arr[i] || pan[arr[i]])

        {

            ans[arr[i]].pop();

            arr[i] = pp.top();

            pp.pop();

            continue;

        }

        

        pan[arr[i]] = true;

        

    }

    cout << sum << endl;

    for(i = 1; i < n; ++i)

        cout << arr[i] << " ";

    cout << arr[i] << endl;

}