CodeForces - 556B Case of Fake Numbers

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作者：stxy-ferryman
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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, ..., a_n (0 ≤ a_i ≤ n - 1) — the sequence of active teeth: the active tooth of the i-th gear contains number a_i.

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.

Example

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.

题意：有n个齿轮，依次相接，齿轮示数为ai，按下某个不可名状的东西以后，奇数齿轮顺时针旋转（示数+1），偶数齿轮逆时针旋转（示数-1），如果能转出0，1，2....n-1则输出yes否则输出no

题解：转一遍，检验一遍，对则输出yes，如果转完n次仍不能成功，那就没有什么价值继续了，输出no

当然，有o(n)算法http://paste.ubuntu.com/11788578/

代码：

var
b:boolean;
n,i,j,k:longint;
a:array[1..1001] of longint;
function jy(n:longint):boolean;
begin
for i:=1 to n do
if a[i]<>i-1 then exit(false);
exit(true);
end;
begin
readln(n);
for i:=1 to n do
read(a[i]);
b:=true;
while (not jy(n)) do
begin
for i:=1 to n do
begin
if i mod 2=1 then inc(a[i])
             else dec(a[i]);
if a[i]<0 then a[i]:=n-1;
if a[i]>n-1 then a[i]:=0;
end;
inc(k);
if k>n then begin b:=false; break; end;
end;
if b then writeln('Yes')
     else writeln('No');
end.