题目大意：

This is an interactive problem. Refer to the Interaction section below for better understanding.

Ithea and Chtholly want to play a game in order to determine who can use the kitchen tonight.

Initially, Ithea puts n clear sheets of paper in a line. They are numbered from 1 to n from left to right.

This game will go on for m rounds. In each round, Ithea will give Chtholly an integer between 1 and c, and Chtholly needs to choose one of the sheets to write down this number (if there is already a number before, she will erase the original one and replace it with the new one).

Chtholly wins if, at any time, all the sheets are filled with a number and the n numbers are in non-decreasing order looking from left to right from sheet 1 to sheet n, and if after m rounds she still doesn't win, she loses the game.

Chtholly really wants to win the game as she wants to cook something for Willem. But she doesn't know how to win the game. So Chtholly finds you, and your task is to write a program to receive numbers that Ithea gives Chtholly and help her make the decision on which sheet of paper write this number.

Input

The first line contains 3 integers n, m and c (,  means  rounded up) — the number of sheets, the number of rounds and the largest possible number Ithea can give to Chtholly respectively. The remaining parts of input are given throughout the interaction process.

Interaction

In each round, your program needs to read one line containing a single integer pi (1 ≤ pi ≤ c), indicating the number given to Chtholly.

Your program should then output a line containing an integer between 1 and n, indicating the number of sheet to write down this number in.

After outputting each line, don't forget to flush the output. For example:

• fflush(stdout) in C/C++;
• System.out.flush() in Java;
• sys.stdout.flush() in Python;
• flush(output) in Pascal;
• See the documentation for other languages.

If Chtholly wins at the end of a round, no more input will become available and your program should terminate normally. It can be shown that under the constraints, it's always possible for Chtholly to win the game.

Example

Input

2 4 4
2
1
3

Output

1
2
2

Note

In the example, Chtholly initially knew there were 2 sheets, 4 rounds and each number was between 1 and 4. She then received a 2 and decided to write it in the 1st sheet. Then she received a 1 and wrote it in the 2nd sheet. At last, she received a 3 and replaced 1 with 3in the 2nd sheet. At this time all the sheets were filled with a number and they were non-decreasing, so she won the game.

Note that it is required that your program terminate immediately after Chtholly wins and do not read numbers from the input for the remaining rounds. If not, undefined behaviour may arise and it won't be sure whether your program will be accepted or rejected. Also because of this, please be careful when hacking others' codes. In the sample, Chtholly won the game after the 3rd round, so it is required that your program doesn't read the number of the remaining 4th round.

The input format for hacking:

• The first line contains 3 integers n, m and c;
• The following m lines each contains an integer between 1 and c, indicating the number given to Chtholly in each round.

题目大意：

给一个有n个格子从左到右排列的白板，系统每次随机给你一个1-c的数字， 请你设计一个算法，每轮把系统给你的这个数字填到白板上，如果白板的这一格已经填有数字，就将其擦去填新的数。要求这个算法在给定的m轮内填满n个格子，并保证这n个格子从左至右是一个不减序列 。每次输出填到哪里。

解题报告：

   注意到m的范围，给了很明显的提示了，这个m还是很大的，相当于n个格子每个都可以写c/2次，那就把可能出现的数字1-c二分，如果小于等于c/2就从左边开始填，否则从右边开始填，那么一定可以构造出来。填的时候左侧维护一个递增序列，右侧维护一个递减序列，如果遇到空的格子直接填上去，做不到维护不增就替换第一个大于（右边填起就是小于）这个数的数字。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
int now;
int n,m,c,a[MAX];
int main()
{cin>>n>>m>>c;c/=2;while(m--) {int x;scanf("%d",&x);if(x <= c) {for(int i = 1; i<=n; i++) {if(a[i] == 0 || a[i] > x) {a[i] = x;printf("%d\n",i);//cout << i << endl;fflush(stdout);break;}}}else {for(int i = n; i>=1; i--) {if(a[i] == 0 || a[i] < x) {a[i] = x;printf("%d\n",i);//cout << i << endl;fflush(stdout);break;}}}now = 0;for(int i = 1; i<=n; i++) {if(a[i] != 0) now++;}if(now == n) break;}return 0 ;}