Spider Man CodeForces

Spider Man

CodeForces - 705B

Peter Parker wants to play a game with Dr. Octopus. The game is about cycles. Cycle is a sequence of vertices, such that first one is connected with the second, second is connected with third and so on, while the last one is connected with the first one again. Cycle may consist of a single isolated vertex.

Initially there are k cycles, i-th of them consisting of exactly v_i vertices. Players play alternatively. Peter goes first. On each turn a player must choose a cycle with at least2 vertices (for example, x vertices) among all available cycles and replace it by two cycles withp and x - p vertices where1 ≤ p < x is chosen by the player. The player who cannot make a move loses the game (and his life!).

Peter wants to test some configurations of initial cycle sets before he actually plays with Dr. Octopus. Initially he has an empty set. In thei-th test he adds a cycle with a_i vertices to the set (this is actually a multiset because it can contain two or more identical cycles). After each test, Peter wants to know that if the players begin the game with the current set of cycles, who wins?

Peter is pretty good at math, but now he asks you to help.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of tests Peter is about to make.

The second line contains n space separated integersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹),i-th of them stands for the number of vertices in the cycle added before thei-th test.

Output

Print the result of all tests in order they are performed. Print 1 if the player who moves first wins or 2 otherwise.

Example

Input

31 2 3

Output

211

Input

51 1 5 1 1

Output

22222

Note

In the first sample test:

In Peter's first test, there's only one cycle with 1 vertex. First player cannot make a move and loses.

In his second test, there's one cycle with 1 vertex and one with2. No one can make a move on the cycle with 1 vertex. First player can replace the second cycle with two cycles of 1 vertex and second player can't make any move and loses.

In his third test, cycles have 1, 2 and 3 vertices. Like last test, no one can make a move on the first cycle. First player can replace the third cycle with one cycle with size1 and one with size 2. Now cycles have1, 1, 2, 2 vertices. Second player's only move is to replace a cycle of size2 with 2 cycles of size1. And cycles are 1, 1, 1, 1, 2. First player replaces the last cycle with 2 cycles with size1 and wins.

In the second sample test:

Having cycles of size 1 is like not having them (because no one can make a move on them).

In Peter's third test: There a cycle of size 5 (others don't matter). First player has two options: replace it with cycles of sizes1 and 4 or 2 and 3.

• If he replaces it with cycles of sizes 1 and 4: Only second cycle matters. Second player will replace it with 2 cycles of sizes 2. First player's only option to replace one of them with two cycles of size1. Second player does the same thing with the other cycle. First player can't make any move and loses.
• If he replaces it with cycles of sizes 2 and 3: Second player will replace the cycle of size 3 with two of sizes1 and 2. Now only cycles with more than one vertex are two cycles of size2. As shown in previous case, with 2 cycles of size 2 second player wins.

So, either way first player loses.

题意：

    博弈游戏，一堆数，可以拆成两堆非空的，所有堆都不能拆，则为输。每次不断加入一个新堆，问在当前所有堆的情况下的输赢状况。

解题：

   因为一个数字n，可以拆的次数为固定的n-1次，故而只要统计全部的可操作次数，看总和的奇偶性即可。

  

（因此每次加入新的数后，开始的顺序是延续上一次的顺序，上一次到哪里了就从谁开始，即次数累加）

code：

#include <iostream>#include <cmath>using namespace std;int main(){    int n;    cin >> n;    int sum = 0;    int i,temp;    for(i = 0; i < n; i++){        cin >> temp;        sum += temp-1;//每个数有n-1次拆分        if(sum&1)//是奇数最后一个肯定是先手            cout << "1" << endl;        else//是偶数说明一人一次正平分，最后一个能拆分的是后手            cout << "2" << endl;    }    return 0;}