Codeforces Round #446 (Div. 2) C. Pride (贪心 数论)

C. Pride

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacent elements from a, say xand y, and replace one of them with gcd(x, y), where gcd denotes the greatest common divisor.

What is the minimum number of operations you need to make all of the elements equal to 1?

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements in the array.

The second line contains n space separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.

Examples

input

52 2 3 4 6

output

5

input

42 4 6 8

output

-1

input

32 6 9

output

4

Note

In the first sample you can turn all numbers to 1 using the following 5 moves:

• [2, 2, 3, 4, 6].
• [2, 1, 3, 4, 6]
• [2, 1, 3, 1, 6]
• [2, 1, 1, 1, 6]
• [1, 1, 1, 1, 6]
• [1, 1, 1, 1, 1]

We can prove that in this case it is not possible to make all numbers one using less than 5 moves.

#include <bits/stdc++.h>using namespace std;int a[2001];int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}int main(){int n;cin >> n;for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);}int cur = a[1];for(int i = 2; i <= n; ++i){cur = gcd(cur, a[i]);}if(cur != 1){cout << -1 << endl;return 0;}int cnt = 0;for(int i = 1; i <= n; ++i){if(a[i] == 1){cnt++;}}int ans = 1e9;for(int i = 1; i <= n; ++i){if(a[i] == 1){cout << n - cnt << endl;return 0;}cur = a[i];for(int j = i; j <= n; ++j){cur = gcd(cur, a[j]);if(cur == 1){ans = min(ans, j - i + 1);break;}}}cout << (ans + n - 2 - cnt) << endl;}/*题意：2000个数，每次操作可以让相邻两个数的其中一个变成两者的最大公约数。问最少需要操作多少次可以让所有数变成1。思路：首先排除无解的情况，所有数的最大公约数不是1，那么一定无解，否则一定有解。如果有数字是1，那么只需要操作不是1的数字个数的次数即可。剩下的情况枚举区间[i,j]，找最小区间GCD为1的，然后维护答案。*/