Ilya lives in a beautiful city of Chordalsk.

There are n houses on the street Ilya lives, they are numerated from 1 to n from left to right; the distance between every two neighboring houses is equal to 1 unit. The neighboring houses are 1 and 2, 2 and 3, ..., n-1 and n. The houses n and 1 are not neighboring.

The houses are colored in colors c_1, c_2, …, c_n so that the i-th house is colored in the color c_i. Everyone knows that Chordalsk is not boring, so there are at least two houses colored in different colors.

Ilya wants to select two houses i and j so that 1 ≤ i < j ≤ n, and they have different colors: c_i ≠ c_j. He will then walk from the house i to the house j the distance of (j-i) units.

Ilya loves long walks, so he wants to choose the houses so that the distance between them is the maximum possible.

Help Ilya, find this maximum possible distance.

Input

The first line contains a single integer n (3 ≤ n ≤ 300 000) — the number of cities on the street.

The second line contains n integers c_1, c_2, …, c_n (1 ≤ c_i ≤ n) — the colors of the houses.

It is guaranteed that there is at least one pair of indices i and j so that 1 ≤ i < j ≤ n and c_i ≠ c_j.

Output

Print a single integer — the maximum possible distance Ilya can walk.

Examples

Input


5
1 2 3 2 3


Output


4


Input


3
1 2 1


Output


1


Input


7
1 1 3 1 1 1 1


Output


4

Note

In the first example the optimal way is to walk from the first house to the last one, where Ilya can walk the distance of 5-1 = 4 units.

In the second example the optimal way is to either walk from the first house to the second or from the second to the third. Both these ways have the distance of 1 unit.

In the third example the optimal way is to walk from the third house to the last one, where Ilya can walk the distance of 7-3 = 4 units. 
"""
Author    : co_devil Chirag Garg
Institute : JIIT
"""
from math import log, ceil
from collections import defaultdict
from math import *
from sys import stdin, stdout
import itertools
import os
import sys
import threading
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import *
# from math import ceil, floor, log, sqrt, factorial, pow, pi, gcd
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
from fractions import Fraction
mod = int(pow(10, 9)+7)


def ii(): return int(input())


def si(): return str(input())


def mi(): return map(int, input().split())


def li(): return list(mi())


def fii(): return int(stdin.readline())


def fsi(): return str(stdin.readline())


def fmi(): return map(int, stdin.readline().split())


def fli(): return list(fmi())


abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
       'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
       'z': 25}


def getKey(item): return item[0]


def sort2(l): return sorted(l, key=getKey)


def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]


def isPowerOfTwo(x): return (x and (not (x & (x - 1))))


def decimalToBinary(n): return bin(n).replace("0b", "")


def ntl(n): return [int(i) for i in str(n)]


def powerMod(x, y, p):
    res = 1
    x %= p
    while y > 0:
        if y & 1:
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res


graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000


def bfs(d, v):
    q = []
    q.append(v)
    visited[v] = 1
    while len(q) != 0:
        x = q[0]
        q.pop(0)
        for i in d[x]:
            if visited[i] != 1:
                visited[i] = 1
                q.append(i)
        print(x)


def make_graph(e):
    d = {}
    for i in range(e):
        x, y = mi()
        if x not in d:
            d[x] = [y]
        else:
            d[x].append(y)
        if y not in d:
            d[y] = [x]
        else:
            d[y].append(x)
    return d


def gr2(n):
    d = defaultdict(list)
    for i in range(n):
        x, y = mi()
        d[x].append(y)
    return d


def connected_components(graph):
    seen = set()

    def dfs(v):
        vs = set([v])
        component = []
        while vs:
            v = vs.pop()
            seen.add(v)
            vs |= set(graph[v]) - seen
            component.append(v)
        return component

    ans = []
    for v in graph:
        if v not in seen:
            d = dfs(v)
            ans.append(d)
    return ans


def primeFactors(n):
    s = set()
    while n % 2 == 0:
        s.add(2)
        n = n // 2
    for i in range(3, int(sqrt(n)) + 1, 2):
        while n % i == 0:
            s.add(i)
            n = n // i
    if n > 2:
        s.add(n)
    return s


def find_all(a_str, sub):
    start = 0
    while True:
        start = a_str.find(sub, start)
        if start == -1:
            return
        yield start
        start += len(sub)


def SieveOfEratosthenes(n, isPrime):
    isPrime[0] = isPrime[1] = False
    for i in range(2, n):
        isPrime[i] = True
    p = 2
    while (p * p <= n):
        if (isPrime[p] == True):
            i = p * p
            while (i <= n):
                isPrime[i] = False
                i += p
        p += 1
    return isPrime


def dijkstra(edges, f, t):
    g = defaultdict(list)
    for l, r, c in edges:
        g[l].append((c, r))

    q, seen, mins = [(0, f, ())], set(), {f: 0}
    while q:
        (cost, v1, path) = heappop(q)
        if v1 not in seen:
            seen.add(v1)
            path = (v1, path)
            if v1 == t:
                return (cost, path)

            for c, v2 in g.get(v1, ()):
                if v2 in seen:
                    continue
                prev = mins.get(v2, None)
                next = cost + c
                if prev is None or next < prev:
                    mins[v2] = next
                    heappush(q, (next, v2, path))
    return float("inf")


def binsearch(a, l, r, x):
    while l <= r:
        mid = l + (r-1)//2
        if a[mid]:
            return mid
        elif a[mid] > x:
            l = mid-1
        else:
            r = mid+1
    return -1


# def dfs(d, v):
#     q = []
#     q.insert(0, v)
#     visited[v] = 1
#     while len(q) != 0:
#         x = q[0]
#         q.pop(0)
#         for i in d[x]:
#             if visited[i] != 1:
#                 visited[i] = 1
#                 q.insert(0, i)
#         print(x)

n = ii()
l = li()
x, y = 0, 0
for i in range(n-1):
    if l[i] != l[-1]:
        x = i
        break
for i in range(n-1, 0, -1):
    if l[i] != l[0]:
        y = i
        break
print(max(n-1-x, y))
