#!/usr/bin/env python3
n = int(input())

ans = [None] * n

def query(a, b):
    print('?', a+1, b+1, flush=True)
    x, y = map(int, input().split())
    return (x,y)

# If querying (i, i+1) and (i+1, i+2) both give the same pair (a,b), position i can be both a and b.
# Store this, and resolve conflicts in the end when we know the rest of the array.

ans[0], ans[1] = query(0, 1)
last_ans = None
for i in range(1, n-1):
    x, y = query(i, i+1)
    if {x,y} == last_ans:
        ans[i-1] = {x, y}

    last_ans = {x, y}
    if ans[i] not in [x,y]:
        ans[i], ans[i-1] = ans[i-1], ans[i]
    assert ans[i] in [x,y]
    ans[i] = x
    assert ans[i+1] is None
    ans[i+1] = y

# Count the number of times each integer occurs, and fix unresolved positions.
count = {i: 0 for i in range(1, n+1)}
for x in ans:
    if isinstance(x, int):
        count[x] += 1

for i, x in enumerate(ans):
    if isinstance(x, set):
        for y in x:
            if count[y] < 2:
                ans[i] = y
                count[y] += 1
                break

# The two remaining unfixed positions
uk = (n-2, n-1)
while ans[uk[0]] != ans[uk[1]]:
    # Find another position with the given value.
    p1 = uk[0]
    p2 = None
    for i in range(n):
        if ans[i] == ans[uk[0]] and i != p1:
            p2 = i
            break
    assert p2 is not None
    x, y = query(p1, p2)

    # Happens with probability 0.5.
    if x == y: break

    z = ans[uk[1]]
    assert x == z or y == z

    ans[uk[0]], ans[uk[1]] = ans[uk[1]], ans[uk[0]]
    uk = (p1, p2)


print('!', *ans)