Codeforce Round #213 Div2 B

B. The Fibonacci Segment

time limit per test 1 second

memory limit per test 256 megabytes

 

You have array a₁, a₂, ..., a_n. Segment [l, r] (1 ≤ l ≤ r ≤ n) is good if a_i = a_i - 1 + a_i - 2, for all i (l + 2 ≤ i ≤ r).

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, ..., a_n (0 ≤ a_i ≤ 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

#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 1000001
#define INF 1000000000
#define ll long long
int a[maxn];
int ans[maxn];
ll gcd(ll n, ll m){ return m ? gcd(m, n%m) : n; }
int n, m, b, c, d, p,t1, t2, cnt1, cnt2, add, sub;
int main(){
    int t, cas = 1;
    scanf("%d", &n);
    int k = 0;
    for (int i = 0; i < n; i++){scanf("%d", &a[i]);}
    for (int i = 2; i < n; i++){
        if (a[i] == a[i - 1] + a[i - 2])ans[k]++;
        else k++;
    }
    m = ans[0];
    for (int i = 0; i <= n; i++)if (m < ans[i])m = ans[i];
    if (n == 1)m = -1;
    printf("%d\n", m+2);
    return 0;
}