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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"sort"
	"strconv"
)

//Util
func nextStr() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func maxInt(a, b int) int {
	if a > b {
		return a
	} else {
		return b
	}
}

func minInt(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}

func absInt(a int) int {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func absFloat64(a float64) float64 {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func maxFloat64(a, b float64) float64 {
	if a > b {
		return a
	} else {
		return b
	}
}

func minFloat64(a, b float64) float64 {
	if a > b {
		return b
	} else {
		return a
	}
}

func str2Int(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}

func reverse(s string) string {
	runes := []rune(s)
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return string(runes)
}

func pow(p, q int) int {
	return int(math.Pow(float64(p), float64(q)))
}

func isPrime(x int) bool {
	if x == 1 {
		return false
	}
	if x == 2 {
		return true
	}
	if x%2 == 0 {
		return false
	}

	b := true
	rootx := int(math.Sqrt(float64(x)))
	i := 3
	for i <= rootx {
		if x%i == 0 {
			b = false
			break
		}
		i += 2
	}
	return b
}

func PrimeFactors(n int) (pfs []int) {
	// Get the number of 2s that divide n
	for n%2 == 0 {
		pfs = append(pfs, 2)
		n = n / 2
	}

	// n must be odd at this point. so we can skip one element
	// (note i = i + 2)
	for i := 3; i*i <= n; i = i + 2 {
		// while i divides n, append i and divide n
		for n%i == 0 {
			pfs = append(pfs, i)
			n = n / i
		}
	}

	// This condition is to handle the case when n is a prime number
	// greater than 2
	if n > 2 {
		pfs = append(pfs, n)
	}

	return
}

//Main
const (
	initialBufSize = 10000
	maxBufSize     = 1000000
	mod            = 1e9 + 7
)

var (
	sc *bufio.Scanner = func() *bufio.Scanner {
		sc := bufio.NewScanner(os.Stdin)
		buf := make([]byte, initialBufSize)
		sc.Buffer(buf, maxBufSize)
		return sc
	}()
)

var N int

func main() {
	sc.Split(bufio.ScanWords)

	N = nextInt()

	A := make([]int, N+1)
	B := make([]int, N+1)
	for i := 1; i <= N; i++ {
		A[i] = nextInt()
		B[i] = A[i]
	}

	sort.Ints(B)
	for i := 1; i <= N; i++ {
		if B[N] != A[i] {
			fmt.Println(B[N])
		} else {
			fmt.Println(B[N-1])
		}
	}
}