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

There are n holds on the wall, i-th hold is at height a_i off the ground. Besides, let the sequence a_i increase, that is, a_i < a_i + 1 for all i from 1 to n - 1; we will call such sequence a track. Mike thinks that the track a₁, ..., a_n has difficulty . 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, ..., a_n. 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.

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

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

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

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.

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 1001

const int inf=0x7fffffff;   //无限大

int h[maxn];

int main()

{

    int n;

    cin>>n;

    cin>>h[];

    int minn=;

    for(int i=;i<n-;i++)

    {

        cin>>h[i];

        minn=max(h[i]-h[i-],minn);

    }

    cin>>h[n-];

    minn=max(h[n-]-h[n-],minn);

    //cout<<minn<<endl;

    int kiss=;

    for(int i=;i<n;i++)

    {

        kiss=min(h[i]-h[i-],kiss);

    }

    //cout<<kiss<<endl;

    cout<<max(kiss,minn)<<endl;

    return ;

}