Mike is trying rock climbing but he is awful at it. 

There are n holds on the wall, i-th hold is at height ai off the ground. Besides, let the sequence ai increase, that is, ai < ai + 1 for all i from 1 to n - 1; we will call such sequence a track. Mike thinks that the track a1, ..., an has difficulty <image>. In other words, difficulty equals the maximum distance between two holds that are adjacent in height.

Today Mike decided to cover the track with holds hanging on heights a1, ..., an. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1, 2, 3, 4, 5) and remove the third element from it, we obtain the sequence (1, 2, 4, 5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.

Help Mike determine the minimum difficulty of the track after removing one hold.

Input

The first line contains a single integer n (3 ≤ n ≤ 100) — the number of holds.

The next line contains n space-separated integers ai (1 ≤ ai ≤ 1000), where ai is the height where the hold number i hangs. The sequence ai is increasing (i.e. each element except for the first one is strictly larger than the previous one).

Output

Print a single number — the minimum difficulty of the track after removing a single hold.

Examples

Input

3
1 4 6


Output

5


Input

5
1 2 3 4 5


Output

2


Input

5
1 2 3 7 8


Output

4

Note

In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.

In the second test after removing every hold the difficulty equals 2.

In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
# pylint: disable=unused-variable
# pylint: enable=too-many-lines
#* Just believe in yourself


#@ Author @CAP
#import numpy
import os
import sys
from io import BytesIO, IOBase
import math as M
import itertools as ITR
from collections import defaultdict as D
from collections import Counter as C
from collections import deque as Q
import threading
from functools import lru_cache, reduce
from functools import cmp_to_key as CMP
from bisect import bisect_left as BL
from bisect import bisect_right as BR
import random as R
import string
import cmath,time
enum=enumerate
start_time = time.time()

#* Variables

MOD=1_00_00_00_007; MA=float("inf"); MI=float("-inf")

#* Graph 8 direction
di8=((1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1))

#* Graph 4 direction
di4=((1, 0), (0, 1), (-1, 0), (0, -1))

#* Stack increment


def increase_stack():
    sys.setrecursionlimit(2**32//2-1)
    threading.stack_size(1 << 27)
    #sys.setrecursionlimit(10**6)
    #threading.stack_size(10**8)
    # t = threading.Thread(target=main)
    # t.start()
    # t.join()


#* Region Funtions


def binary(n): return bin(n)[2:]


def decimal(s): return (int(s, 2))


def pow2(n):
    p = 0
    while (n > 1):
        n //= 2
        p += 1
    return (p)

def maxfactor(n):
    q = []
    for i in range(1,int(n ** 0.5) + 1):
        if n % i == 0: q.append(i)
    if q:
        return q[-1]


def factors(n):
    q = []
    for i in range(1,int(n ** 0.5) + 1):
        if n % i == 0: q.append(i); q.append(n // i)
    return(list(sorted(list(set(q)))))

def primeFactors(n):
    l=[]
    while n % 2 == 0:
        l.append(2)
        n = n / 2
    for i in range(3, int(M.sqrt(n)) + 1, 2):
        while n % i == 0:
            l.append(i)
            n = n / i
    if n > 2:
        l.append(int(n))
    l.sort()
    return (l)


def isPrime(n):
    if (n == 1):
        return (False)
    else:
        root = int(n ** 0.5)
        root += 1
        for i in range(2, root):
            if (n % i == 0):
                return (False)
        return (True)


def seive(n):
    a = []
    prime = [True for i in range(n + 1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p ** 2,n + 1,p): 
                prime[i] = False
        p = p + 1
    for p in range(2,n + 1): 
        if prime[p]: 
            a.append(p)
    prime[0]=prime[1]=False
    return a,prime


def maxPrimeFactors(n):
    maxPrime = -1
    while n % 2 == 0:
        maxPrime = 2
        n >>= 1
    for i in range(3, int(M.sqrt(n)) + 1, 2):
        while n % i == 0:
            maxPrime = i
            n = n / i
    if n > 2:
        maxPrime = n
    return int(maxPrime)


def countchar(s,i):
    c=0
    ch=s[i]
    for i in range(i,len(s)):
        if(s[i]==ch):
            c+=1
        else:
            break
    return(c)

def str_counter(a):
    q = [0] * 26
    for i in range(len(a)):
        q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
    return(q)

def lis(arr):
    n = len(arr)
    lis = [1] * n
    maximum=0

    for i in range(1, n):
        for j in range(0, i):
            if arr[i] > arr[j] and lis[i] < lis[j] + 1:
                lis[i] = lis[j] + 1
                maximum=max(maximum,lis[i])
    return maximum


def lcm(arr):
    a=arr[0]; val=arr[0]

    for i in range(1,len(arr)):

        gcd=gcd(a,arr[i])
        a=arr[i]; val*=arr[i]

    return val//gcd

def ncr(n,r): return M.factorial(n) // (M.factorial(n - r) * M.factorial(r))


def npr(n,r): return M.factorial(n) // M.factorial(n - r)




#* Make work easy funtions

def IF(c, t, f): return t if c else f

def YES(c): print(IF(c, "YES", "NO"))

def Yes(c): print(IF(c, "Yes", "No"))

def yes(c): print(IF(c, "yes", "no"))

def JA(a, sep=" "): print(sep.join(map(str, a)))

def JAA(a, s='\n', t=" "): return s.join(t.join(map(str, b)) for b in a)

def PS(a,s=" "): print(str(a),end=s) 

#* Region Taking Input

def I(): return int(inp())

def F(): return float(inp())

def LI(): return list(map(int,inp().split()))

def LF(): return list(map(float,inp().split()))

def MATI(n): return [ LI() for i in range(n)]

def MATS(n): return [list(inp()) for i in range(n)]

def IV(): return map(int,inp().split())

def FV(): return map(float,inp().split())

def LS(): return list(inp())

def S(): return inp()


#* Region Fastio

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)
inp = lambda: sys.stdin.readline().rstrip("\r\n")


#<==================================== Write The Useful Code Here ============================

#< Make it one if there is some test cases
TestCases=0      #<=====================
#< =======================================
"""
> Sometimes later becomes never. Do it now.
! Be Better than yesterday.
* Your limitation—it’s only your imagination.
> Push yourself, because no one else is going to do it for you.
? The harder you work for something, the greater you’ll feel when you achieve it.
! Great things never come from comfort zones.
* Don’t stop when you’re tired. Stop when you’re done.
> Do something today that your future self will thank you for.
? It’s going to be hard, but hard does not mean impossible.
! Sometimes we’re tested not to show our weaknesses, but to discover our strengths.
"""
#@  Goal is to get Candidate Master



def solve():
    n = I()
    nums = LI()

    ma = MI; mi = MA

    for i in range(1, n):
        ma = max(ma, nums[i]-nums[i-1])

    for i in range(2, n):
        val = max(ma, nums[i]-nums[i-2])
        mi = min(mi, val)

    print(mi)


#* This is the Main Function

def main():
    flag=1
    #! Checking we are offline or not
    try:
        sys.stdin = open("c:/Users/Manoj Chowdary/Documents/python/CodeForces/contest-div-2/input.txt","r")
        sys.stdout = open("c:/Users/Manoj Chowdary/Documents/python/CodeForces/contest-div-2/output.txt","w")
    except: flag=0

    t=1
    if TestCases:
        t = I()
    for _ in range(1,t + 1):
        solve()

    if flag:
        print("Time: %.4f sec"%(time.time() - start_time))
        localtime = time.asctime( time.localtime(time.time()) )
        print(localtime)
        sys.stdout.close()



#* End Region


if __name__ == "__main__":

    #? Incresing Stack Limit
    #increase_stack()

    #! Calling Main Function
    main()