You are given a permutation of 1,2,...,N: p_1,p_2,...,p_N. Determine if the state where p_i=i for every i can be reached by performing the following operation any number of times:
 - Choose three elements p_{i-1},p_{i},p_{i+1} (2\leq i\leq N-1) such that p_{i-1}>p_{i}>p_{i+1} and reverse the order of these three.

-----Constraints-----
 - 3 \leq N \leq 3 × 10^5
 - p_1,p_2,...,p_N is a permutation of 1,2,...,N.

-----Input-----
Input is given from Standard Input in the following format:
N
p_1
:
p_N

-----Output-----
If the state where p_i=i for every i can be reached by performing the operation, print Yes; otherwise, print No.

-----Sample Input-----
5
5
2
1
4
3

-----Sample Output-----
Yes

The state where p_i=i for every i can be reached as follows:
 - Reverse the order of p_1,p_2,p_3. The sequence p becomes 1,2,5,4,3.
 - Reverse the order of p_3,p_4,p_5. The sequence p becomes 1,2,3,4,5.

Difficulty: competition

Starter code:

Solutions:
import sys


def solve(ppp):
    section_start = -1
    moved_left_max = 0
    moved_right_max = 0
    prev = True

    for i, p in enumerate(ppp, start=1):
        if i == p:
            if prev:
                moved_left_max = 0
                moved_right_max = 0
                section_start = -1
            prev = True
        else:
            if not prev:
                if moved_left_max > i - 1:
                    return False

                moved_left_max = 0
                moved_right_max = 0
                section_start = i

            if section_start == -1:
                section_start = i

            if i > p:
                if section_start > p:
                    return False
                if moved_right_max > p:
                    return False
                moved_right_max = p
            else:
                if moved_left_max > p:
                    return False
                moved_left_max = p

            prev = False
    return True


n, *ppp = list(map(int, sys.stdin))
print(('Yes' if solve(ppp) else 'No'))


def Split(a):
    no = []
    for i, x in a:
        if no and (i == x) == (no[-1][0] == no[-1][1]):
            yield no
            no = []
        no.append((i, x))
    yield no

for sq in Split((i + 1, int(input())) for i in range(int(input()))):
    tb = [0, 0]
    for np, goal in sq:
        if goal != np:
            if goal < tb[np < goal] or goal > sq[-1][0] or goal < sq[0][0]:
                print("No")
                return
            tb[np < goal] = goal
print("Yes")

def Split(a):
    no = []
    for i, x in a:
        if no:
            is_ok = i == x
            la_ok = no[-1][0] == no[-1][1]
            if is_ok == la_ok:
                yield no
                no = []
        no.append((i, x))
    yield no

n = int(input())
p = list(enumerate((int(input()) for i in range(n)), 1))
for sq in Split(p):
    tl = tr = 0
    for np, goal in sq:
        if goal > np:
            if goal < tr or goal > sq[-1][0]:
                print("No")
                return
            tr = goal
        elif goal < np:
            if goal < tl or goal < sq[0][0]:
                print("No")
                return
            tl = goal
print("Yes")