All cinema halls in Berland are rectangles with K rows of K seats each, and K is an odd number. Rows and seats are numbered from 1 to K. For safety reasons people, who come to the box office to buy tickets, are not allowed to choose seats themselves. Formerly the choice was made by a cashier, but now this is the responsibility of a special seating program. It was found out that the large majority of Berland's inhabitants go to the cinema in order to watch a movie, that's why they want to sit as close to the hall center as possible. Moreover, a company of M people, who come to watch a movie, want necessarily to occupy M successive seats in one row. Let's formulate the algorithm, according to which the program chooses seats and sells tickets. As the request for M seats comes, the program should determine the row number x and the segment [y_l, y_r] of the seats numbers in this row, where y_r - y_l + 1 = M. From all such possible variants as a final result the program should choose the one with the minimum function value of total seats remoteness from the center. Say, $\text{[math]}$ — the row and the seat numbers of the most "central" seat. Then the function value of seats remoteness from the hall center is $\text{[math]}$ . If the amount of minimum function values is more than one, the program should choose the one that is closer to the screen (i.e. the row number x is lower). If the variants are still multiple, it should choose the one with the minimum y_l. If you did not get yet, your task is to simulate the work of this program.

Input

The first line contains two integers N and K (1 ≤ N ≤ 1000, 1 ≤ K ≤ 99) — the amount of requests and the hall size respectively. The second line contains N space-separated integers M_i from the range [1, K] — requests to the program.

Output

Output N lines. In the i-th line output «-1» (without quotes), if it is impossible to find M_i successive seats in one row, otherwise output three numbers x, y_l, y_r. Separate the numbers with a space.

Idx	Name	Rating
A	Power Consumption Calculation	900
B	Cinema Cashier	1500
C	Digital Root	2000
D	LCIS	2800
E	Greedy Change	2600

Codeforces Beta Round 10