CF#213DIV2：B The Fibonacci Segment

Let's define len([l, r]) = r - l + 1, len([l, r]) is the length of the segment [l, r]. Segment [l1, r1], is longer than segment [l2, r2], if len([l1, r1]) > len([l2, r2]).

Your task is to find a good segment of the maximum length in array a. Note that a segment of length 1 or 2 is always good.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of elements in the array. The second line contains integers: a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

Print the length of the longest good segment in array a.

Sample test(s)

Input

10

1 2 3 5 8 13 21 34 55 89

Output

10

Input

5

1 1 1 1 1

Output

2

找出符合a[i] = a[i-1]+a[i-2]的最长长度

没有考虑0的状况。。。。。。蛋疼，长久没做CF了，坑成球了。。。

#include   
#include   
#include   
using namespace std;  
  
int a[100005],b[100005],len;  
  
int main()  
{  
    int maxn,sum;  
    int i,n;  
    while(~scanf("%d",&n))  
    {  
        sum = 0;  
        for(i = 0; i





maxn && r!=l)  
                    maxn = r-l+1;  
                flag = 1;  
                l = 0;  
                r = 0;  
            }  
        }  
        if(r-l+1>maxn && r!=l)  
            maxn = r-l+1;  
        printf("%d\n",maxn+2);  
    }  
  
    return 0;  
}