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.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;



public class Main {
	 
    public static void main(String[] args) {
    	
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        solve(in, out);
        out.close();
    }
 
    static String reverse(String s) {
        return (new StringBuilder(s)).reverse().toString();
    }

    static void sieveOfEratosthenes(int n, int factors[]) 
    { 
       n=1000000001;
    	factors[1]=1;
        for(int p = 2; p*p <=n; p++) 
        { 
            if(factors[p] == 0) 
            { 
                factors[p]=p;
                for(int i = p*p; i <= n; i += p) 
                    factors[i] = p; 
            } 
        } 
    }

    static void sort(int ar[]) {
        int n = ar.length;
        ArrayList<Integer> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static void sort1(long ar[]) {
        int n = ar.length;
        ArrayList<Long> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static long ncr(long n, long r, long mod) {
        if (r == 0)
            return 1;
        long val = ncr(n - 1, r - 1, mod);
        val = (n * val) % mod;
        val = (val * modInverse(r, mod)) % mod;
        return val;
    }
    public static int convert(String a, String b)
    {
    	int T=0;
    	if(b.charAt(0)=='P')
    	{
    		if(a.charAt(0)=='1'&&a.charAt(1)=='2') T=12;
    		else
    			T=(a.charAt(0)-'0')*10+(a.charAt(1)-'0')+12;
    	}
    	else if(b.charAt(0)=='A')
    	{
    		if(a.charAt(0)=='1'&&a.charAt(1)=='2') T=0;
    		else
    			T=(a.charAt(0)-'0')*10+(a.charAt(1)-'0');
    	}
    		
    	T=T*100+(a.charAt(3)-'0')*10+(a.charAt(4)-'0');
    	return T;
    	
    }
    static boolean areBracketsBalanced(String expr) 
    { 
        // Using ArrayDeque is faster than using Stack class 
        Deque<Character> stack 
            = new ArrayDeque<Character>(); 
  
        // Traversing the Expression 
        for (int i = 0; i < expr.length(); i++)  
        { 
            char x = expr.charAt(i); 
  
            if (x == '(' || x == '[' || x == '{')  
            { 
                // Push the element in the stack 
                stack.push(x); 
                continue; 
            } 
  
            // IF current current character is not opening 
            // bracket, then it must be closing. So stack 
            // cannot be empty at this point. 
            if (stack.isEmpty()) 
                return false; 
            char check; 
            switch (x) { 
            case ')': 
                check = stack.pop(); 
                if (check == '{' || check == '[') 
                    return false; 
                break; 
  
            case '}': 
                check = stack.pop(); 
                if (check == '(' || check == '[') 
                    return false; 
                break; 
  
            case ']': 
                check = stack.pop(); 
                if (check == '(' || check == '{') 
                    return false; 
                break; 
            } 
        } 
  
        // Check Empty Stack 
        return (stack.isEmpty()); 
    } 
    public static int hr(String s)
    {
    	int hh=0;
    	hh=(s.charAt(0)-'0')*10+(s.charAt(1)-'0');
    	return hh;
    }
    public static int mn(String s)
    {
    	int mm=0;
    	mm=(s.charAt(3)-'0')*10+(s.charAt(4)-'0');
    	return mm;
    }
	
	/*
	 * public static void countNumberOfPrimeFactors(){ boolean flag[]=new
	 * boolean[N]; Arrays.fill(flag,true); for(int i=2;i<N;i++){ if(flag[i]){
	 * for(int j=i;j<N;j+=i) { prime[j]++; flag[j]=false; } } } // prime[1]=1; }
	 */	 

   

	
	public static String rev(String s)
	{
		char[] temp=s.toCharArray();
		for(int i=0;i<s.length()/2;i++)
		{
			char tmp=temp[i];
			temp[i]=temp[s.length()-1-i];
			temp[s.length()-1-i]=tmp;
			
		}
		String str="";
		for(char ch: temp)
		{
			str+=ch;
		}
		return str;
	}
	static boolean isP(String str)
    {
 
        // Pointers pointing to the beginning
        // and the end of the string
        int i = 0, j = str.length() - 1;
 
        // While there are characters to compare
        while (i < j) {
 
            // If there is a mismatch
            if (str.charAt(i) != str.charAt(j))
                return false;
 
            // Increment first pointer and
            // decrement the other
            i++;
            j--;
        }
 
        // Given string is a palindrome
        return true;
    }
	public static boolean isSorted(List<Integer> l)
	{
		int n=l.size(),i=0;
		for(i=1;i<l.size();i++)
		{
			if(l.get(i-1)<l.get(i)) return false;
		}
		return true;
	}

public static boolean ln(long n, int c)
{
	int l=0;
	long t=n;
	while(t!=0)
	{
		
		t/=10;
		l++;
	}
	return l==c;
}
	
  
    public static void solve(InputReader sc, PrintWriter pw) {
  //  	int t=sc.nextInt();
  //   	 while(t-->0) {
     		 int n=sc.nextInt();
     		 int a[]=new int[n];
     		 for(int i=0;i<n;i++)
     		 {
     			a[i]=sc.nextInt(); 
     		 }
     	    if(n==3) {pw.print(a[n-1]-a[0]);}
     	    else
     	    {
     	    	int mx=0;
     	    	int smx=0;
     	    	for(int i=0;i<n-1;i++)
     	    	{
     	    		mx=Math.max(a[i+1]-a[i],mx);
     	    	}
     	   	int smn=Integer.MAX_VALUE;
     	    	for(int i=0;i<n-2;i++)
     	    	{
     	    	
     	    		if(smn>a[i+2]-a[i]) smn=a[i+2]-a[i];
     	    	
     	      
     	        	   
     	    	}
     	    	
     	    	pw.print(smn>=mx?smn:mx);
     	    	
     	    }
     		
     	// }	
        
    }
    static class Pair1 {
        long a;
        long b;
        
        Pair1(long a, long b) {
            this.a = a;
            this.b = b;
        }
    }

    static class Pair implements Comparable<Pair> {
        int a;
        int b;
        
        Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
 
        public int compareTo(Pair p) {
            if(a!=p.a)
                return p.a-a;
            return b-p.b;
        }
    }

    static boolean isPrime(long n) {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
        if (n % 2 == 0 || n % 3 == 0)
            return false;
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        return true;
    }

    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static long fast_pow(long base, long n, long M) {
        if (n == 0)
            return 1;
        if (n == 1)
            return base % M;
        long halfn = fast_pow(base, n / 2, M);
        if (n % 2 == 0)
            return (halfn * halfn) % M;
        else
            return (((halfn * halfn) % M) * base) % M;
    }

    static long modInverse(long n, long M) {
        return fast_pow(n, M - 2, M);
    }
    public static int LowerBound(int a[], int x) 
	   {
			  int l=-1,r=a.length;
			  while(l+1<r) {
			    int m=(l+r)>>>1;
			    if(a[m]>=x) r=m;
			    else l=m;
			  }
			  return r;
		}
	
	   
	public static int UpperBound(int a[], int x) 
		{
		    int l=-1, r=a.length;
		    while(l+1<r) {
		       int m=(l+r)>>>1;
		       if(a[m]<=x) l=m;
		       else r=m;
		    }
		    return l+1;
		}

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }
    }
}


