Problem:
Score : 500 points

Problem Statement

You are going to hold a competition of one-to-one game called AtCoder Janken. (Janken is the Japanese name for Rock-paper-scissors.)
N players will participate in this competition, and they are given distinct integers from 1 through N.
The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive).
You cannot assign the same integer to multiple playing fields.
The competition consists of N rounds, each of which proceeds as follows:

For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there.

Then, each player adds 1 to its integer. If it becomes N+1, change it to 1.

You want to ensure that no player fights the same opponent more than once during the N rounds.
Print an assignment of integers to the playing fields satisfying this condition.
It can be proved that such an assignment always exists under the constraints given.

Constraints

1 \leq M

M \times 2 +1 \leq N \leq 200000

Input

Input is given from Standard Input in the following format:

N M

Output

Print M lines in the format below.
The i-th line should contain the two integers a_i and b_i assigned to the i-th playing field.

a_1 b_1
a_2 b_2
:
a_M b_M

Sample Input 1

4 1

Sample Output 1

2 3

Let us call the four players A, B, C, and D, and assume that they are initially given the integers 1, 2, 3, and 4, respectively.

The 1-st round is fought by B and C, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 2, 3, 4, and 1, respectively.

The 2-nd round is fought by A and B, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 3, 4, 1, and 2, respectively.

The 3-rd round is fought by D and A, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 4, 1, 2, and 3, respectively.

The 4-th round is fought by C and D, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 1, 2, 3, and 4, respectively.

No player fights the same opponent more than once during the four rounds, so this solution will be accepted.

Sample Input 2

7 3

Sample Output 2

1 6
2 5
3 4

Solution in JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

function Scanner(input) {
	var self = this;
	self.itemNo = 0;
	self.orgStr = input;
	self.splitedStrs = (function(target) {
		var result = [];
		var target = target.replace(/[\n \t]+/g, " ");
		var strs = target.split(" ");
		for (var i = 0, len = strs.length; i < len; i++) {
			result.push(strs[i]);
		}
		return result;
	}(input));

	self.next = function() {
		var result = self.splitedStrs[self.itemNo];
		self.itemNo++;
		return (result != null) ? result : null;
	}

	self.nextInt = function() {
		var result = self.splitedStrs[self.itemNo];
		self.itemNo++;
		return (result != null) ? parseInt(result) : null;
	}

	self.nextBigDecimal = function() {
		var result = self.splitedStrs[self.itemNo];
		self.itemNo++;
		return (result != null) ? parseFloat(result) : null;
	}
	
	self.nextByte = function() {
		var result = self.splitedStrs[self.itemNo];
		self.itemNo++;
		return (result != null) ? new Int8Array(result) : null;
	}

}

function Main(sc) {

	var N = sc.nextInt();
	var M = sc.nextInt();
	var anss = [];
	
	if (N == M * 2 + 1) {
		for (var i = 0; i < M; i++) {
			var ans = (Math.floor(N / 2) - i) + " " + (Math.floor(N / 2) + i + 1);
			anss.push(ans);
		}
	} else {
		for (var i = 0; i < M; i++) {
			var start = 0;
			var skip = 0;
			if (i % 2 == 0) {
				start = parseInt(i / 2) + 1;
				skip = M - i;
			} else {
				start = parseInt(i / 2) + parseInt(N / 2) + 1;
				skip = M - i;
			}
			var ans = start + " " + (start + skip);
			anss.push(ans);
		}
	}
	
	console.log(anss.join("\n"));
}
Main(new Scanner(require("fs").readFileSync("/dev/stdin", "utf8")));
// Main(new Scanner("33"));

// var str1 = sc.next(); // Stringで取得
// var int1 = sc.nextInt(); // intで取得
// var bd1 = sc.nextBigDecimal(); // BigDecimalで取得
// var byte1 = sc.nextByte(); // byteで取得
// for (var i = 0; i < 1000000000; i++) var toDelay = "aaa"; // 1000msくらいの遅延