Problem:
Score : 300 points

Problem Statement

You are given a sequence of length N: A_1, A_2, ..., A_N.
For each integer i between 1 and N (inclusive), answer the following question:

Find the maximum value among the N-1 elements other than A_i in the sequence.

Constraints

2 \leq N \leq 200000

1 \leq A_i \leq 200000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print N lines. The i-th line (1 \leq i \leq N) should contain the maximum value among the N-1 elements other than A_i in the sequence.

Sample Input 1

3
1
4
3

Sample Output 1

4
3
4

The maximum value among the two elements other than A_1, that is, A_2 = 4 and A_3 = 3, is 4.

The maximum value among the two elements other than A_2, that is, A_1 = 1 and A_3 = 3, is 3.

The maximum value among the two elements other than A_3, that is, A_1 = 1 and A_2 = 4, is 4.

Sample Input 2

2
5
5

Sample Output 2

5
5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#define start_routine() int begtime = clock();
#define end_routine() int endtime = clock(); cerr << endl << "Time elapsed: " << (endtime - begtime)*1000/CLOCKS_PER_SEC << " ms"; return 0
#define speed() cin.tie(0), cout.tie(0), ios_base::sync_with_stdio(false)
// #define exit(a, b) return cout << a, b;

#define loop(i,a,b) for(ll i=a;i<b;i++)
#define all(v) v.begin(), v.end() 

#define print(stuff) cout << stuff << endl
#define printc(stuff) for(auto x: stuff) cout << x << " "; cout << endl;
#define printPrec(stuff) cout << fixed << setprecision(15) << stuff << endl;
#define debug(stuff) cout << #stuff << ": " << stuff <<endl
#define debugc(stuff) cout << #stuff << ": "; for(auto x: stuff) cout << x << " "; cout << endl;
#define len length
#define ret0 return 0
#define ret return 

#define ll long long
#define ld long double
#define fi first
#define endl '\n'
#define se second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define fill(ar, x) memset(ar, x, sizeof ar)

#define vl vector<ll> 
#define sl set<ll>
#define pll pair<ll, ll>
#define mll map<ll, ll> 
#define pq priority_queue<ll>

// typedef tree<ll,null_type,less<ll>,rb_tree_tag,
// tree_order_statistics_node_update> indexed_set;

#define inf (long long int) 1e18
#define eps 0.000001
#define mod 1000000007
#define mod1 998244353
#define MAXN (ll)1e5+5

void solve(){
    ll n;
    cin>>n;
    ll a[n];
    ll b[n];
    loop(i,0,n){
        cin>>a[i];
        b[i]=a[i];
    }
    sort(b, b+n);
    ll mx1=b[n-1];
    ll mx2=b[n-2];
    loop(i,0,n){
        if(a[i]!=mx1){
            print(mx1);
        }
        else if(a[i]==mx1){
            print(mx2);
        }
    }
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    

    ll t = 1;
    // cin>>t;
    while(t--){
        solve();
    }
        
   
}