time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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 the i-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.

Examples

input

3

1 0 0

output

Yes

input

5

4 2 1 4 3

output

Yes

input

4

0 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.

【题目链接】:

【题解】

当英语的阅读题能吓死不少人。。。

读懂题意后其实很简单了；

第一个齿轮它的活跃齿肯定是0;看看第一个齿轮要转几次;

则其他齿轮也只能转相应的次数；

如果转了相应的次数某个齿轮i不能变成i-1;则NO;

【完整代码】

#include 
     
      using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%I64d",&x)#define pri(x) printf("%d",x)#define prl(x) printf("%I64d",x)typedef pair
      
        pii;typedef pair
       
         pll;const int MAXN = 1e3+10;const int dx[9] = {
    0,1,-1,0,0,-1,-1,1,1};const int dy[9] = {
    0,0,0,-1,1,-1,1,-1,1};const double pi = acos(-1.0);int n;int a[MAXN];void cl(int tag,int &t){    if (tag==1)    {        t++;        if (t>n-1) t = 0;    }    else    {        t--;        if (t<0) t = n-1;    }}int main(){    //freopen("F:\\rush.txt","r",stdin);    rei(n);    rep1(i,1,n)        rei(a[i]);    int now = 1;    int quan = 0;    while (a[1]!=0)        {             cl(now,a[1]);             quan++;        }    rep1(i,2,n)    {        now*=-1;        int cnt = 0;        while (a[i]!=i-1)        {            cl(now,a[i]);            cnt++;            if (cnt>quan)            {                puts("No");                return 0;            }        }        if (cnt < quan)        {            puts("No");            return 0;        }    }    puts("Yes");    return 0;}