# 2023. iCow

Fatigued by the endless toils of farming, Farmer John has decided to try his hand in the MP3 player market with the new iCow. It is an MP3 player that stores N songs (1 <= N <= 1,000) indexed 1 through N that plays songs in a “shuffled” order, as determined by Farmer

John’s own algorithm:

• Each song i has an initial rating R_i (1 <= R_i <= 10,000).

• The next song to be played is always the one withthe highest rating (or, if two or more are tied, the highest rated song with the lowest index is chosen).

• After being played, a song’s rating is set to zero, and its rating points are distributed evenly among the other N-1 songs.

• If the rating points cannot be distributed evenly (i.e.,they are not divisible by N-1), then the extra points are parceled out one at a time to the first songs on the list (i.e., R_1, R_2, etc. – but not the played song) until no more extra points remain.

This process is repeated with the new ratings after the next songis played.

Determine the first T songs (1 <= T <= 1000) that are played by theiCow.

### 输入格式

• Line 1: Two space-separated integers: N and T

• Lines 2..N+1: Line i+1 contains a single integer: R_i

### 输出格式

• Lines 1..T: Line i contains a single integer that is the i-th song that the iCow plays.

### 样例

Input
3 4
10
8
11
INPUT DETAILS:
The iCow contains 3 songs, with ratings 10, 8, and 11, respectively.You must determine the first 4 songs to be played.

Output
3
1
2
3
OUTPUT DETAILS:
The ratings before each song played are:
R_1 R_2 R_3
10 8 11 -> play #3 11/2 = 5, leftover = 1
16 13 0 -> play #1 16/2 = 8
0 21 8 -> play #2 21/2 = 10, leftover = 1
11 0 18 -> play #3 ...


40 人解决，52 人已尝试。

43 份提交通过，共有 156 份提交。

4.3 EMB 奖励。