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;

#define int long long
#define MOD 1000000007

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep1(i, n) for (int i = 1; i < (n); i++)
#define rrep(i, n) for (int i = (n)-1; i >= 0; i--)
#define all(i) (i).begin(), (i).end()
#define rall(i) (i).begin(), (i).end(), greater<int>()

#define elif else if
#define eb emplace_back
#define pb push_back
#define mp make_pair
#define fst first
#define sec second
template <typename T>
void print(T x)
{
    std::cout << x << '\n';
}

//typedef long long ll;
#define TENNINE 1000000005
#define TENFIVE 100005
//#define int long long

int N;
vector<int> a;

signed main()
{
    cin >> N;
    a.reserve(N);

    rep(i, N)
    {
        cin >> a[i];
    }

    int max_l[N];
    int max_r[N];

    max_l[0] = a[0];
    max_r[N - 1] = a[N - 1];

    for (int i = 1; i < N; i++)
    {
        max_l[i] = max(a[i], max_l[i - 1]);
        max_r[N - 1 - i] = max(a[N - 1 - i], max_r[N - i]);
    }

    cout << max_r[1] << endl;

    for (int i = 1; i < N - 1; i++)
    {
        cout << max(max_l[i - 1], max_r[i + 1]) << endl;
    }

    cout << max_l[N - 2] << endl;

    return 0;
}