There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.

At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?

Constraints

* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N


Output

Print the minimum possible number of different characters contained in the strings.

Examples

Input

3
3 2 1


Output

2


Input

5
2 3 2 1 2


Output

2
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <bitset>
using namespace std;
const int maxn=200010;

int n,top,a[maxn];
struct node{
	int w,a;
}zhan[maxn];
bool bz;

void doit(int p,int q)
{
	if (p==0)
	{
		bz=false;
		return;
	}
	while (zhan[top].w>p)
	{
		top--;
	}
	if (zhan[top].w==p) zhan[top].a++;
	else 
	{
		top++;
		zhan[top].w=p;
		zhan[top].a=2;
	}
	if (zhan[top].a==q+1)
	{
		if (top>0)
		{
			top--;
			doit(p-1,q);
		}
	}
}

bool pd(int x)
{
	top=0;zhan[0].w=0;
	bz=true;
	for (int i=1;i<=n;i++)
	{
		if (a[i]<=a[i-1])
		{
			doit(a[i],x);
		}
	}
	return bz;
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	
	int l=2;
	int r=100000000;
	int ans=0;
	while (l<=r)
	{
		int mid=(l+r)/2;
		if (pd(mid))
		{
			ans=mid;
			r=mid-1;
		}
		else
		{
			l=mid+1;
		}
	}
	int k=0;
	for (int i=1;i<n;i++)
	{
		if (a[i]>=a[i+1])
		{
			k=1;
		}
	}
	if (k==0) ans=1;
	printf("%d\n",ans);
}