Codeforces Round #348 (VK Cup 2016 Round 2, Div. 2 Edition) D

D. Little Artem and Dance

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Artem is fond of dancing. Most of all dances Artem likes rueda — Cuban dance that is danced by pairs of boys and girls forming a circle and dancing together.

More detailed, there are n pairs of boys and girls standing in a circle. Initially, boy number 1 dances with a girl number 1, boy number 2 dances with a girl number 2 and so on. Girls are numbered in the clockwise order. During the dance different moves are announced and all pairs perform this moves. While performing moves boys move along the circle, while girls always stay at their initial position. For the purpose of this problem we consider two different types of moves:

1. Value x and some direction are announced, and all boys move x positions in the corresponding direction.
2. Boys dancing with even-indexed girls swap positions with boys who are dancing with odd-indexed girls. That is the one who was dancing with the girl 1 swaps with the one who was dancing with the girl number 2, while the one who was dancing with girl number 3 swaps with the one who was dancing with the girl number 4 and so one. It's guaranteed that n is even.

Your task is to determine the final position of each boy.

Input

The first line of the input contains two integers n and q (2 ≤ n ≤ 1 000 000, 1 ≤ q ≤ 2 000 000) — the number of couples in the rueda and the number of commands to perform, respectively. It's guaranteed that n is even.

Next q lines contain the descriptions of the commands. Each command has type as the integer 1 or 2 first. Command of the first type is given as x ( - n ≤ x ≤ n), where 0 ≤ x ≤ n means all boys moves x girls in clockwise direction, while  - x means all boys move x positions in counter-clockwise direction. There is no other input for commands of the second type.

Output

Output n integers, the i-th of them should be equal to the index of boy the i-th girl is dancing with after performing all q moves.

Examples

Input

6 3
1 2
2
1 2

Output

4 3 6 5 2 1

Input

2 3
1 1
2
1 -2

Output

1 2

Input

4 2
2
1 3

Output

1 4 3 2

题意：1~n编号的女孩与1~n编号的男孩配对围圈跳舞
     两种操作
     1 x 若x为正值 所有的男孩顺时针移动x个位置 继续与女孩跳舞 若为负值 则逆时针移动
     2 奇偶位置男孩交换位置（这里的奇偶位置是 根据初始的参考位置）
     q个操作后 输出与1~n编号的女孩跳舞的男孩的编号序列
题解：

        模拟发现 奇数或偶数 位置的移动量都是相同的；

        只需要模拟出1，2位置的移动量jj，ou； 注意取%

        遍历得出所有序列；

我gou 码

  #include<iostream>
  #include<cstring>
  #include<cstdio>
  #include<queue>
  #include<stack>
  #include<map>
  #include<set>
  #include<algorithm>
  #define ll __int64
  #define pi acos(-1.0)
  #define mod 1
  #define maxn 10000
  using namespace std;
 int n,qq;
 int ans[];
 int mo(int qqq)
 {return (qqq+*n)%n;}
 int main()
 {
     int jj,ou,ke,k,gg;
     scanf("%d %d",&n,&qq);
     jj=;ou=;
     for(int i=;i<=qq;i++)
     {
         scanf("%d",&k);
         if(k==)
         {
              scanf("%d",&gg);
              jj+=gg;
              jj=mo(jj);
              ou+=gg;
              ou=mo(ou);
         }
         else
         {
             if((jj+n)%)
             {
                 jj--;
                 jj=mo(jj);
                 ou++;
                 ou=mo(ou);
             }
             else
             {
                 jj++;
                 jj=mo(jj);
                 ou--;
                 ou=mo(ou);
             }
         }

     }
     for(int i=;i<=n;i++)
     {
         if(i%){
             ke=i+jj;
             ke=mo(ke);
             ans[ke]=i;
         }
         else{
             ke=i+ou;
             ke=mo(ke);
             ans[ke]=i;
         }
     }
     for(int i=;i<n;i++)
         printf("%d ",ans[i]);
     printf("%d\n",ans[]);
     return ;
 }