B. Case of Fake Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Andrewid the Android is a galaxy-famous detective. He is now investigating a case of frauds who make fake copies of the famous Stolp's gears, puzzles that are as famous as the Rubik's cube once was.

Its most important components are a button and a line of n similar gears. Each gear has n teeth containing all numbers from 0 to n - 1 in the counter-clockwise order. When you push a button, the first gear rotates clockwise, then the second gear rotates counter-clockwise, the the third gear rotates clockwise an so on.

Besides, each gear has exactly one active tooth. When a gear turns, a new active tooth is the one following after the current active tooth according to the direction of the rotation. For example, if n = 5, and the active tooth is the one containing number 0, then clockwise rotation makes the tooth with number 1 active, or the counter-clockwise rotating makes the tooth number 4 active.

Andrewid remembers that the real puzzle has the following property: you can push the button multiple times in such a way that in the end the numbers on the active teeth of the gears from first to last form sequence 0, 1, 2, ..., n - 1. Write a program that determines whether the given puzzle is real or fake.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of gears.

The second line contains n digits a1, a2, ..., an (0 ≤ ai ≤ n - 1) — the sequence of active teeth: the active tooth of thei-th gear contains number ai.

Output

In a single line print "Yes" (without the quotes), if the given Stolp's gears puzzle is real, and "No" (without the quotes) otherwise.

Sample test(s)

input

31 0 0

output

Yes

input

54 2 1 4 3

output

Yes

input

40 2 3 1

output

No

Note

In the first sample test when you push the button for the first time, the sequence of active teeth will be 2 2 1, when you push it for the second time, you get 0 1 2.

解题说明：题目的意思是有n个数字转盘，每个转盘都是由0 ~ n - 1这n个数字按逆时针排列，每个转盘都有一个活动数字。现在有个按钮，每按一次按钮，每个转盘的活动数字会按如下规则变化：第奇数个转盘的活动数字变为之前活动数字按逆时针方向的下一个数字（即 + 1）。第偶数个转盘的活动数字变为之前活动数字按顺时针方向的下一个数字（即 - 1）。问是否可以通过按若干次按钮，使得第 i 个转盘的活动数字为i。显然按动n次按钮就会还原。所以次数在0 ~ n - 1之间。而显然在这些次数中，按不同次数的按钮得到的每个转盘的活动数字都不会一样（因为恰好是n个数字），所以直接计算将第一个转盘的活动数字变为0需要多少次，然后按照这个次数去转其他的转盘看是否能满足第 i 个转盘的数字恰好为 i 即可。

#include<stdio.h>#include <string.h>int main(){int n,a[1005],i,r;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);}r=n-a[0];for(i=1;i<n;i++){if(i%2==1){a[i]=a[i]-r;}else{a[i]+=r;}a[i]+=n;a[i]%=n;if(a[i]!=i){break;}}if(i==n){printf("Yes\n");}else{printf("No\n");}return 0;}