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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <tuple>
#include <cstdio>
#include <bitset>
#include <sstream>
#include <iterator>
#include <numeric>
#include <map>
#include <cstring>
#include <set>
#include <functional>
#include <iomanip>

using namespace std;

#define DEBUG_ //!!提出時にコメントアウト!!
#ifdef DEBUG_
	#define dump(x)  cerr << #x << " = " << (x) << endl;
#else
	#define dump(x)  ;
#endif

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define SZ(x) ((int)(x).size())
#define pb push_back
#define eb emplace_back

//#define int long long

typedef long long LL;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

template <typename T>
std::string printVector(const std::vector<T> &data)
{
    std::stringstream ss;
    std::ostream_iterator<T> out_it(ss, ", ");
    ss << "[";
    std::copy(data.begin(), data.end() - 1, out_it);
    ss << data.back() << "]";
    return ss.str();
}

const int MOD = 1e9+7;
const LL LINF = 1001002003004005006ll;
const int INF = 1001001001;

signed main(int argc, char const *argv[])
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; cin >> N;
    VL A(N);
    REP(i,N){
        cin >> A[i];
    }
    VL B(N),C(N);

    B[0] = A[0];
    C[N-1] = A[N-1];
    FOR(i,1,N){
        B[i] = max(B[i-1],A[i]);
    }
    for(int i = N-2; i >= 0; i--){
        C[i] = max(C[i+1],A[i]);
    }
    LL ans;
    REP(i,N){
        if(i == 0) ans = C[1];
        else if(i == N-1) ans = B[N-2];
        else{
            ans = max(B[i-1],C[i+1]);
        }
        cout << ans << endl;
    }

}