## solution

$n \equiv 0 \pmod{4}$ であることと目的の対が存在することは同値。 対の位置を見付けるのはいい感じに二分探索。 二分探索。

## note

$n \equiv 0 \pmod{4}$ の条件は実験で出したので未証明。 二分探索が常に上手くいくことも未証明。 editorialまだですか。

## implementation

#!/usr/bin/env python3
import sys
print('?', i + 1)
sys.stdout.flush()
a_i = int(input())
return a_i
print('!', i + 1 if i != -1 else -1)
sys.exit()

def has_intersection(l1, r1, l2, r2):
if l1 <= l2 and r2 <= r1:
return True
if l2 <= l1 and r1 <= r2:
return True
return False

n = int(input())
assert n >= 2 and n % 2 == 0
if (n // 2) % 2 == 1:
else:
assert n % 4 == 0
l1 = 0
r1 = n // 2
if a_l1 == a_r1:
a_l2 = a_r1
a_r2 = a_l1
# print('binary search [', l1, ',', r1, ') ->', (l1 + r1) // 2, file=sys.stderr)
while True:
m1 = (l1 + r1) // 2
m2 = (m1 + n // 2) % n
if a_m1 == a_m2:
if has_intersection(a_l1, a_m1, a_l2, a_m2):
r1 = m1
a_r1 = a_m1
a_r2 = a_m2
else:
assert has_intersection(a_m1, a_r1, a_m2, a_r2)
l1 = m1
a_l1 = a_m1
a_l2 = a_m2
# print('binary search [', l1, ',', r1, ') ->', (l1 + r1) // 2, file=sys.stderr)
assert False

#!/usr/bin/env python3
import sys
import random
a = [ 1 ]
while True:
if a[-1] >= -10 and random.random() < 0.5 + a[-1] / 100:
a += [ a[-1] - 1 ]
else:
a += [ a[-1] + 1 ]
if a[-1] == 0 and len(a) >= 60 and len(a) % 4 == 0:
break
k = random.randrange(len(a))
a = a[k :] + a[: k]
# a = [1, 0, -1, -2, -1, 0, 1, 2, 3, 2, 1, 0, 1, 2, 3, 2]
# a = [3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 4, 3, 2]
# a = a = [8, 9, 8, 7, 6, 5, 6, 5, 6, 5, 6, 7, 6, 7, 6, 5, 4, 3, 2, 3, 2, 1, 2, 1, 0, 1, 2, 3, 2, 1, 0, 1, 0, 1, 2, 3, 2, 3, 2, 3, 4, 5, 6, 5, 6, 7, 8, 9, 8, 7, 6, 7]
n = len(a)
print('[*] n =', n, file=sys.stderr)
print('[*] a =', a, file=sys.stderr)

print(len(a))
sys.stdout.flush()
for _ in range(61):
c, i = input().split()
print('[>]', c, i, file=sys.stderr)
i = int(i) - 1
if c == '?':
print(a[i])
sys.stdout.flush()
print('[<]', a[i], file=sys.stderr)
elif c == '!':
if i == -2:
for j in range(n):
assert a[j] != a[(j + n // 2) % n]
else:
assert a[i] == a[(i + n // 2) % n]
break
else:
assert False
else:
assert False