CodeForces1159BExpansion coefficient of the array

B. Expansion coefficient of the array

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Let's call an array of non-negative integers a1,a2,…,an a k-extension for some non-negative integer k if for all possible pairs of indices 1≤i,j≤n the inequality k⋅|i−j|≤min(ai,aj) is satisfied. The expansion coefficient of the array a is the maximal integer k such that the array a is a k-extension. Any array is a 0-expansion, so the expansion coefficient always exists.

You are given an array of non-negative integers a1,a2,…,an. Find its expansion coefficient.

Input

The first line contains one positive integer n — the number of elements in the array a (2≤n≤300000). The next line contains n non-negative integers a1,a2,…,an, separated by spaces (0≤ai≤10^9).

Output

Print one non-negative integer — expansion coefficient of the array a1,a2,…,an.

Examples

input

4

6 4 5 5

output

1

 

input

3

0 1 2

output

0

 

input

4

821 500 479 717

output

239

 

Note

In the first test, the expansion coefficient of the array [6,4,5,5] is equal to 1 because |i−j|≤min(ai,aj), because all elements of the array satisfy ai≥3. On the other hand, this array isn't a 2-extension, because 6=2⋅|1−4|≤min(a1,a4)=5 is false.

In the second test, the expansion coefficient of the array [0,1,2] is equal to 0 because this array is not a 1-extension, but it is 0-extension.

题解

对于数列 a1,a2,…,an，分析其中任意一项ai，考虑所有的aj >= ai（aj < ai的算在aj里面考虑了），则使得min(ai,aj) = ai，要使所有的j都满足k*|i-j| <= min(ai,aj) = ai，则对于ai来说，j应该尽量远离i，这样解出来的最大的k才是有用的（即对ai的k值的约束最紧）。显然最远的j应该是数列两端中的一端，但两端不一定大于ai。那怎么办呢？仔细一想，对于两端的情况，考虑最前端a1及最后端an：

1.  如果a1 >= ai，则对于当前ai来说，k要满足k <= ai/|i-1|（如果i == 1，则说明k值没有限制）。如果a1 < ai，则对于当前ai来说，k要满足k <= a1/|i-1| < ai/|i-1|。
2.  如果an >= ai，则对于当前ai来说，k要满足k <= ai/|i-n|（如果i == n，则说明k值没有限制）。如果an < ai，则对于当前ai来说，k要满足k <= an/|i-n| < ai/|i-n|。
3.  在所有大于等于ai的aj中，k对于ai来说要满足k <= ai/|i-j|，而|i-j| <= |i-1|或|i-j| <= |i-n|，故ai/|i-1| <= ai/|i-j|或ai/|i-n| <= ai/|i-j|。

可见，对每项ai来说，考虑两端的项所解出来的k值必定有一项是有用的（即对ai的k值的约束最紧）。

枚举每一项求解出来的有用的k值，再取最小即得到最优答案。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string.h>
#include <algorithm>
#define re register
#define il inline
#define ll long long
#define ld long double
const ll MAXN = 1e6+5;
const ll INF = 1e9;

//快读
il ll read()
{
    char ch = getchar();
    ll res = 0, f = 1;
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')   f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        res = (res<<1) + (res<<3) + (ch-'0');
        ch = getchar();
    }
    return res*f;
}

ll a[MAXN];

int main()
{
    ll n = read();
    for(re ll i = 1; i <= n; ++i)
    {
        a[i] = read();
    }
    ll k = INF;
    for(re ll i = 1; i <= n; ++i)
    {
        k = std::min(i==1?k:std::min(a[i],a[1])/(i-1),k);
        k = std::min(i==n?k:std::min(a[i],a[n])/(n-i),k);
    }
    printf("%lld\n", k);
    return 0;
}