D. Little Artem and Dance

题目链接：http://codeforces.com/problemset/problem/669/D

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:

Value x and some direction are announced, and all boys move x positions in the corresponding direction.
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

Copy

Output

Copy

4 3 6 5 2 1

Input

Copy

Output

Copy

1 2

Input

Copy

4 2
2
1 3

Output

Copy

1 4 3 2

题目大意：输入n,代表有n对男女，1···n。所有人围成一个圈， 刚开始男1与女1，男2与女2···对应，q代表有q次操作，如果输入的是1，则所有男的移动x位，正代表顺时针，负代表逆时针。如果输入的是2，则男1与男2交换位子，男3与
男4交换位子···
个人思路：不论怎么操作，其实男1后面两位一定是3，后面4位一定是5·····男2后面两位一定是4，后面四位一定是6····，所以我们只要记录男1和男2的位子，就能知道所有人的位子了
看代码（但这题要用scanf，不然会超时）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
#include<map>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const int maxn=1e6+10;
const int maxk=100+10;
const int maxx=1e4+10;
const ll maxa=43200;
#define INF 0x3f3f3f3f3f3f
int a[maxn];
int main()
{
    int n,q,p1,p2,m,x;
    scanf("%d%d",&n,&q);
    p1=0;
    p2=1;
    for(int i=1;i<=q;i++)
    {
        scanf("%d",&m);
        if(m==1)
        {
            scanf("%d",&x);
            if(x<0)
                x+=n;
            p1+=x;
            p2+=x;
            p1%=n;
            p2%=n;
        }
        else
        {
            if(p1%2==0)
            {
                p1+=1;
            }
            else
            {
                p1-=1;
                //p1=(p1+n)%n;
            }
            if(p2%2==0)
            {
                p2+=1;
            }
            else
            {
                p2-=1;
              //  p2=(p2+6)%6;
            }
        }
    }
    a[p1]=1;
    a[p2]=2;
    int t=n/2;
    int sum=2;
    while(--t)
    {
        sum+=2;
        a[(p2+2)%n]=sum;
        p2=(p2+2)%n;
    }
    t=n-n/2;
    sum=1;
    while(--t)
    {
        sum+=2;
        a[(p1+2)%n]=sum;
        p1=(p1+2)%n;
    }
    for(int i=0;i<n;i++)
        printf("%d ",a[i]);
    printf("\n");
    return 0;
}

D. Little Artem and Dance

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