codeforce 436 D贪心思维题Make a Permutation!

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 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.

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, ..., a_n (1 ≤ a_i ≤ 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 

http://www.cnblogs.com/GraceSkyer/p/7591345.html

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=2e5+;
int vis[N];
bool used[N];
int p[N],num[N];
struct node
{
    int pos,val;
    bool operator < (const node &A)const
    {
        return pos<A.pos;
    }
} q[N];
int ptos[N];
int main()
{
    int n,tot1=,tot2=,x;
    scanf("%d",&n);
    for(int i=; i<=n; ++i)
    {
        scanf("%d",&x);
        num[i]=x;
        if(!vis[x]) vis[x]=;
        else if(vis[x]==)
        {
            vis[x]=;
            q[tot1].val=x;
            q[tot1++].pos=ptos[x];
            q[tot1].val=x;
            q[tot1++].pos=i;
        }
        else
        {
            ++vis[x];
            q[tot1].val=x;
            q[tot1++].pos=i;
        }
        ptos[x]=i;
    }
    for(int i=; i<=n; ++i) if(!vis[i]) p[tot2++]=i;
    sort(q,q+tot1);
    int ct=;
    for(int i=;i<tot1&&ct<tot2;++i) {
        if(!used[q[i].val]&&q[i].val<p[ct]) used[q[i].val]=;
        else if(vis[q[i].val]!=) {--vis[q[i].val]; num[q[i].pos]=p[ct++];}
    }
    printf("%d\n",tot2);
    for(int i=; i<=n; ++i) printf("%d ",num[i]);
    puts("");
}