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 Rust (1.15.1) (judge verdict: Accepted)

// Original: https://github.com/tanakh/competitive-rs
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
        }
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

mod util {
    #[allow(dead_code)]
    pub fn chmin<T>(x: &mut T, y: T) -> bool
    where
        T: PartialOrd + Copy,
    {
        *x > y && {
            *x = y;
            true
        }
    }

    #[allow(dead_code)]
    pub fn chmax<T>(x: &mut T, y: T) -> bool
    where
        T: PartialOrd + Copy,
    {
        *x < y && {
            *x = y;
            true
        }
    }

    /// 整数除算切り上げ
    #[allow(dead_code)]
    pub fn roundup(a: i64, b: i64) -> i64 {
        (a + b - 1) / b
    }

    #[allow(dead_code)]
    pub fn ctoi(c: char) -> i64 {
        c as i64 - 48
    }
}

#[allow(unused_imports)]
use util::*;

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, VecDeque};

fn main() {
    input! {
        n:usize,
        mut ass:[i64;n]
    }

    // 左右からの累積maxでの解法
    // 1-indexedで考える。便宜的にass[0] = ass[n] = -1 とする。
    let n = n + 1;
    ass.insert(0, -1);
    ass.push(-1);
    // ls[i] = [0..i]までの最大値
    let mut ls = vec![-1; n + 1];
    for i in 1..n {
        ls[i] = max(ls[i - 1], ass[i]);
    }
    // rs[i] = [i+1..n]までの最大値
    let mut rs = vec![-1; n + 1];
    for i in (0..n).rev() {
        rs[i] = max(rs[i + 1], ass[i]);
    }

    for i in 1..n {
        // 答えは ass[i]を除いた時の最大値なので、ls[i-1] or rs[i+1]
        let ans = max(ls[i - 1], rs[i + 1]);
        println!("{}", ans);
    }
}