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Problem J4: Party Invitation

2014 · J4 · Original problem · DMOJ

Problem Description

You are hosting a party and do not have room to invite all of your friends. You use the following unemotional mathematical method to determine which friends to invite.

Number your friends 1,2,…,K1,2,\ldots,K and place them in a list in this order. Then perform mm rounds. In each round, use a number to determine which friends to remove from the ordered list.

The rounds will use numbers r1,r2,…,rmr_1, r_2,\ldots, r_m. In round ii remove all the remaining people in positions that are multiples of rir_i (that is, ri,2ri,3ri,…r_i, 2r_i, 3r_i, \ldots) The beginning of the list is position 1.

Output the numbers of the friends that remain after this removal process.

Input Specification

The first line of input contains the integer K (1≤K≤100)K~(1 \leq K \leq 100). The second line of input contains the integer m (1≤m≤10)m~(1 \leq m \leq 10), which is the number of rounds of removal. The next mm lines each contain one integer. The iith of these lines (1≤i≤m)(1 \leq i \leq m) contains ri (2≤ri≤100)r_i~(2 \leq r_i \leq 100) indicating that every person at a position which is multiple of rir_i should be removed.

Output Specification

The output is the integers assigned to friends who were not removed. One integer is printed per line in increasing sorted order.

Sample Input

10
2
2
3

Output for Sample Input

1
3
7
9

Explanation of Output for Sample Input

Initially, our list of invitees is 1,2,3,4,5,6,7,8,9,101,2,3,4,5,6,7,8,9,10. There will be two rounds of removals. After the first round of removals, we remove the even positions (i.e., every second position), which causes our list of invitees to be 1,3,5,7,91,3,5,7,9. After the second round of removals, we remove every 3rd remaining invitee: thus, we keep 11 and 33, remove 55 and keep 77 and 99, which leaves us with an invitee list of 1,3,7,91,3,7,9.