Tolik and His Uncle
Contest: Codeforces Round 569 (Div. 1)
Difficulty rating: 1800
Tags: constructive algorithms

This morning Tolik has understood that while he was sleeping he had invented an incredible problem which will be a perfect fit for Codeforces! But, as a "Discuss tasks" project hasn't been born yet (in English, well), he decides to test a problem and asks his uncle.

After a long time thinking, Tolik's uncle hasn't any ideas on how to solve it. But, he doesn't want to tell Tolik about his inability to solve it, so he hasn't found anything better than asking you how to solve this task.

In this task you are given a cell field $$$n \cdot m$$$, consisting of $$$n$$$ rows and $$$m$$$ columns, where point's coordinates $$$(x, y)$$$ mean it is situated in the $$$x$$$-th row and $$$y$$$-th column, considering numeration from one ($$$1 \leq x \leq n, 1 \leq y \leq m$$$). Initially, you stand in the cell $$$(1, 1)$$$. Every move you can jump from cell $$$(x, y)$$$, which you stand in, by any non-zero vector $$$(dx, dy)$$$, thus you will stand in the $$$(x+dx, y+dy)$$$ cell. Obviously, you can't leave the field, but also there is one more important condition — you're not allowed to use one vector twice. Your task is to visit each cell of the field exactly once (the initial cell is considered as already visited).

Tolik's uncle is a very respectful person. Help him to solve this task!

Input format:
The first and only line contains two positive integers $$$n, m$$$ ($$$1 \leq n \cdot m \leq 10^{6}$$$) — the number of rows and columns of the field respectively.

Output format:
Print "-1" (without quotes) if it is impossible to visit every cell exactly once.

Else print $$$n \cdot m$$$ pairs of integers, $$$i$$$-th from them should contain two integers $$$x_i, y_i$$$ ($$$1 \leq x_i \leq n, 1 \leq y_i \leq m$$$) — cells of the field in order of visiting, so that all of them are distinct and vectors of jumps between them are distinct too.

Notice that the first cell should have $$$(1, 1)$$$ coordinates, according to the statement.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
The vectors from the first example in the order of making jumps are $$$(0, 2), (0, -1), (1, 0), (0, 1), (0, -2)$$$.

Example input:
2 3

Example output:
1 1
1 3
1 2
2 2
2 3
2 1

Example input:
1 1

Example output:
1 1

Editorial:
First, we are going to describe how to bypass $$$1 \cdot m$$$ strip.
This algorithm is pretty easy — $$$(1, 1)$$$ -> $$$(1, m)$$$ -> $$$(1, 2)$$$ -> $$$(1, m-1)$$$ -> $$$\ldots$$$. Obviously all jumps have different vectors because their lengths are different.
It turns out that the algorithm for $$$n \cdot m$$$ grid is almost the same. Initially, we are going to bypass two uttermost horizontals almost the same way as above — $$$(1, 1)$$$ -> $$$(n, m)$$$ -> $$$(1, 2)$$$ -> $$$(n, m-1)$$$ -> $$$\ldots$$$ -> $$$(1, m)$$$ -> $$$(n, 1)$$$. One can realize that all vectors are different because they have different $$$dy$$$. Note that all of them have $$$|dx| = n-1$$$. Then we will jump to $$$(2, 1)$$$ (using $$$(-(n-2), 0)$$$ vector). Now we have a smaller task for $$$(n-2) \cdot m$$$ grid. One can see that we used only vectors with $$$|dx| \geq n-2$$$, so they don't influence now at all. So the task is fully brought down to a smaller one.