Swaps in Permutation

You are given a permutation of the numbers 1, 2, ..., n and m pairs of positions (aj, bj).

At each step you can choose a pair from the given positions and swap the numbers in that positions. What is the lexicographically maximal permutation one can get?

Let p and q be two permutations of the numbers 1, 2, ..., n. p is lexicographically smaller than the q if a number 1 ≤ i ≤ n exists, sopk = qk for 1 ≤ k < i and pi < qi.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 106) — the length of the permutation p and the number of pairs of positions.

The second line contains n distinct integers pi (1 ≤ pi ≤ n) — the elements of the permutation p.

Each of the last m lines contains two integers (aj, bj) (1 ≤ aj, bj ≤ n) — the pairs of positions to swap. Note that you are given apositions, not the values to swap.

Output

Print the only line with n distinct integers p‘i (1 ≤ p‘i ≤ n) — the lexicographically maximal permutation one can get.

Example

input

9 61 2 3 4 5 6 7 8 91 44 72 55 83 66 9

output

7 8 9 4 5 6 1 2 3分析：一个联通体内的数从大到小排列就好；代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e6+10;
const int mod=1e6+3;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,a[maxn],vis[maxn];
vi b[maxn],c;
set<int>p,q;
void dfs(int now)
{
    vis[now]=1;q.insert(now);c.pb(a[now]);
    for(int x:b[now])if(!vis[x])dfs(x);
}
int main()
{
    int i,j,k,t;
    scanf("%d%d",&n,&m);
    rep(i,1,n)scanf("%d",&a[i]);
    while(m--)
    {
        scanf("%d%d",&j,&k);b[j].pb(k),b[k].pb(j);
        p.insert(j),p.insert(k);
    }
    for(int x:p)
    {
        if(!vis[x])
        {
            q.clear();c.clear();
            dfs(x);
            sort(c.rbegin(),c.rend());j=0;
            for(int y:q)a[y]=c[j++];
        }
    }
    rep(i,1,n)printf("%d ",a[i]);
    //system ("pause");
    return 0;
}