Problem: Vasya had an array of n integers, each element of the array was from 1 to n. He chose m pairs of different positions and wrote them down to a sheet of paper. Then Vasya compared the elements at these positions, and wrote down the results of the comparisons to another sheet of paper. For each pair he wrote either "greater", "less", or "equal".

After several years, he has found the first sheet of paper, but he couldn't find the second one. Also he doesn't remember the array he had. In particular, he doesn't remember if the array had equal elements. He has told this sad story to his informatics teacher Dr Helen.

She told him that it could be the case that even if Vasya finds his second sheet, he would still not be able to find out whether the array had two equal elements. 

Now Vasya wants to find two arrays of integers, each of length n. All elements of the first array must be distinct, and there must be two equal elements in the second array. For each pair of positions Vasya wrote at the first sheet of paper, the result of the comparison must be the same for the corresponding elements of the first array, and the corresponding elements of the second array. 

Help Vasya find two such arrays of length n, or find out that there are no such arrays for his sets of pairs.

Input

The first line of input contains two integers n, m — the number of elements in the array and number of comparisons made by Vasya (1 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).

Each of the following m lines contains two integers a_i, b_i — the positions of the i-th comparison (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i). It's guaranteed that any unordered pair is given in the input at most once.

Output

The first line of output must contain "YES" if there exist two arrays, such that the results of comparisons would be the same, and all numbers in the first one are distinct, and the second one contains two equal numbers. Otherwise it must contain "NO".

If the arrays exist, the second line must contain the array of distinct integers, the third line must contain the array, that contains at least one pair of equal elements. Elements of the arrays must be integers from 1 to n.

Examples

Input


1 0


Output


NO


Input


3 1
1 2


Output


YES
1 3 2 
1 3 1 


Input


4 3
1 2
1 3
2 4


Output


YES
1 3 4 2 
1 3 4 1 

Solution (in PYTHON3):

Solution:
# SHRi GANESHA    author: Kunal Verma #
import os
import sys
from bisect import bisect_left, bisect_right

from collections import Counter, defaultdict
from functools import reduce

from io import BytesIO, IOBase
from itertools import combinations
from math import gcd, inf, sqrt, ceil, floor


#sys.setrecursionlimit(2*10**5)
def lcm(a, b):
    return (a * b) // gcd(a, b)


'''
mod = 10 ** 9 + 7
fac = [1]
for i in range(1, 2 * 10 ** 5 + 1):
    fac.append((fac[-1] * i) % mod)
fac_in = [pow(fac[-1], mod - 2, mod)]
for i in range(2 * 10 ** 5, 0, -1):
    fac_in.append((fac_in[-1] * i) % mod)
fac_in.reverse()


def comb(a, b):
    if a < b:
        return 0
    return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''

MAXN = 1000004
spf = [0 for i in range(MAXN)]
def sieve():
    spf[1] = 1
    for i in range(2, MAXN):
        spf[i] = i
    for i in range(4, MAXN, 2):
        spf[i] = 2

    for i in range(3, ceil(sqrt(MAXN))):
        if (spf[i] == i):
            for j in range(i * i, MAXN, i):
                if (spf[j] == j):
                    spf[j] = i


def getFactorization(x):
    ret = Counter()
    while (x != 1):
        ret[spf[x]] += 1
        x = x // spf[x]
    return ret


def printDivisors(n):
    i = 2
    z = [1, n]
    while i <= sqrt(n):
        if (n % i == 0):
            if (n / i == i):
                z.append(i)
            else:
                z.append(i)
                z.append(n // i)
        i = i + 1
    return z


def create(n, x, f):
    pq = len(bin(n)[2:])
    if f == 0:
        tt = min
    else:
        tt = max
    dp = [[inf] * n for _ in range(pq)]
    dp[0] = x
    for i in range(1, pq):
        for j in range(n - (1 << i) + 1):
            dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
    return dp


def enquiry(l, r, dp, f):
    if l > r:
        return inf if not f else -inf
    if f == 1:
        tt = max
    else:
        tt = min
    pq1 = len(bin(r - l + 1)[2:]) - 1
    return tt(dp[pq1][l], dp[pq1][r - (1 << pq1) + 1])


def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
    x = []
    for i in range(2, n + 1):
        if prime[i]:
            x.append(i)
    return x
def main():
    from collections import defaultdict
    import sys
    n, m = map(int, input().split())
    c = defaultdict(int)
    ans = [i for i in range(1, n + 1)]
    ori = [i for i in range(1, n + 1)]
    s = set()
    ma = defaultdict(int)
    con = defaultdict(list)
    for i in range(m):
        a, b = map(int, input().split())
        c[a] += 1
        c[b] += 1
        ma[a] = max(ma[a], b)
        ma[b] = max(ma[b], a)
        con[a].append(b)
        con[b].append(a)
        s.add((a, b))
    e = -1
    if m == (n * (n - 1)) // 2:
        print("NO")
        sys.exit(0)
    for i in range(1, n + 1):
        if (i, i + 1) not in s and (i + 1, i) not in s and i + 1 <= n:
            e = 1
            ans[i + 1 - 1] = ans[i - 1]
            break
        if (i, i - 1) not in s and (i - 1, i) not in s and i - 1 >= 1:
            e = 1
            ans[i - 1 - 1] = ans[i - 1]
            break
    if e == -1:
        for i in range(1, n + 1):
            if e == 1:
                break
            if c[i] == 0:
                e = 1
                ans[i - 1] = 1
                break
            for j in range(i + 1, n + 1):
                if (i, j) not in s and (j, i) not in s:
                    e = 1
                    ori[0] = ori[i - 1]
                    ori[1] = ori[j - 1]
                    ori[i - 1] = 1
                    ori[j - 1] = 2
                    ans[0] = ans[i - 1]
                    ans[1] = ans[j - 1]
                    ans[i - 1] = 1
                    ans[j - 1] = 1
                    break
        if e == -1:
            print("NO")
            sys.exit(0)
    print("YES")
    print(*ori)
    print(*ans)





# Fast IO Region
BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

if __name__ == '__main__':
    main()