C. Ehab and Prefix MEXs

Given an array aa of length nn, find another array, bb, of length nn such that:

• for each ii (1≤i≤n)(1≤i≤n) MEX({b1MEX({b1, b2b2, ……, bi})=aibi})=ai.

The MEXMEX of a set of integers is the smallest non-negative integer that doesn't belong to this set.

If such array doesn't exist, determine this.

Input

The first line contains an integer nn (1≤n≤1051≤n≤105) — the length of the array aa.

The second line contains nn integers a1a1, a2a2, ……, anan (0≤ai≤i0≤ai≤i) — the elements of the array aa. It's guaranteed that ai≤ai+1ai≤ai+1 for 1≤i<n1≤i<n.

Output

If there's no such array, print a single line containing −1−1.

Otherwise, print a single line containing nn integers b1b1, b2b2, ……, bnbn (0≤bi≤1060≤bi≤106)

If there are multiple answers, print any.

Examples

input

Copy

3
1 2 3

output

Copy

0 1 2 

input

Copy

4
0 0 0 2

output

Copy

1 3 4 0 

input

Copy

3
1 1 3

output

Copy

0 2 1 

Note

In the second test case, other answers like [1,1,1,0][1,1,1,0], for example, are valid.

 

 

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod =1e9+7;
const int N = 1e5+5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}      
int n;
int a[N], st[N], d[N], m;
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i], st[a[i]] = 1;
    for (int i = 1; i <= n; i++)
        if (!st[i])
            d[++m] = i;//用d存下所有缝隙的数
    int j = 0;
    for (int i = 1; i <= n; i++)
    {
        if (a[i] == a[i - 1])
            cout << d[++j] << " ";
        else
            cout << a[i - 1] << " ";
    }
    return 0;
}