题干：

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 qchanges.

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.

题目大意：

给出n的数的序列，这n个数范围为1~n。 现在问最少改变几个数能使这n个数成为1~n的某个全排列，若有多种情况，要求使全排列的字典序最小。输出改变的数的个数和这个全排列。

解题报告：

涉及字典序最小的问题其实比较好构造，就是优先贪心前面的，因为前面的变小了要优秀于后面无论多花里胡哨的操作。

首先需要改变的数很好求出，然后对于输出全排列，就先把需要添加的数求出来，排个序，然后从前到后对原序列贪心搞一搞就好了。vis数组考虑这个数是否出现过。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n,a[MAX],cnt[MAX],res[MAX],tot,ans,vis[MAX];
int main()
{cin>>n;for(int i = 1; i<=n; i++) scanf("%d",a+i),cnt[a[i]]++;for(int i = 1; i<=n; i++) {if(cnt[i] == 0) res[++tot] = i;if(cnt[i] > 1) ans += cnt[i]-1;}sort(res+1,res+tot+1);int cur = 1;for(int i = 1; i<=n; i++) {if(cnt[a[i]] == 1) continue;if(vis[a[i]]) cnt[a[i]]--,a[i] = res[cur++];else {if(res[cur] < a[i]) cnt[a[i]]--,a[i] = res[cur++];else vis[a[i]]=1;}}printf("%d\n",ans);for(int i = 1; i<=n; i++) printf("%d ",a[i]);return 0 ;
}