Problem Description

Peter has a sequence a1,a2,...,an and he define a function on the sequence -- F(a1,a2,...,an)=(f1,f2,...,fn), where fi is the length of the longest increasing subsequence ending with ai.

Peter would like to find another sequence b1,b2,...,bn in such a manner that F(a1,a2,...,an) equals to F(b1,b2,...,bn). Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one.

The sequence a1,a2,...,an is lexicographically smaller than sequence b1,b2,...,bn, if there is such number i from 1 to n, that ak=bk for 1≤k<i and ai<bi.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first contains an integer n (1≤n≤100000) -- the length of the sequence. The second line contains n integers a1,a2,...,an (1≤ai≤109).

 

Output

For each test case, output n integers b1,b2,...,bn (1≤bi≤109) denoting the lexicographically smallest sequence.

 

Sample Input

 

Sample Output

#include<iostream>

#include<stdio.h>

#include<string.h>

#include<algorithm>

using namespace std;

const int N=+;

int a[N],b[N],c[N],ans[N];

int n,cnt;

int lowbit(int x)

{

    return x&(-x);

}

int query(int x)

{

    int res=;

    while(x)

    {

        res=max(c[x],res);//找出前面比他小的数中拥有最长上升子序列的

        x-=lowbit(x);

    }

    return res;

}

void update(int p,int x)

{

    while(p<=cnt)

    {

        c[p]=max(x,c[p]);//把后面的数最长上升子序列进行更新

        p+=lowbit(p);

    }

}

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

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

        {

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

            b[i]=a[i];

        }

        sort(b+,b++n);

        cnt=unique(b+,b++n)-b-;

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

        {

            a[i]=lower_bound(b+,b++cnt,a[i])-b;

        }

        memset(c,,sizeof(c));

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

        {

            ans[i]=query(a[i]-)+;//前面存在的最长上升子序列长度加上1，就是当前数拥有的最长上升子序列长度

            update(a[i],ans[i]);//更新后面的数的长度

            printf("%d%c",ans[i],i==n?'\n':' ');

        }

    }

    return ;

}