Codeforces Round #564 (Div. 2)B. Nauuo and Chess

Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m×m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game’s goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (ri,ci), while the following rule is satisfied: for all pairs of pieces i and j, |ri−rj|+|ci−cj|≥|i−j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn’t find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input
The only line contains a single integer n (1≤n≤1000) — the number of chess pieces for the game.

Output
The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers ri and ci (1≤ri,ci≤m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples
inputCopy
2
outputCopy
2
1 1
1 2
inputCopy
4
outputCopy
3
1 1
1 3
3 1
3 3
Note
In the first example, you can’t place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

In the second example, you can’t place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

then |r1−r3|+|c1−c3|=|1−2|+|1−1|=1, |1−3|=2, 1<2; and |r1−r4|+|c1−c4|=|1−2|+|1−2|=2, |1−4|=3, 2<3. It doesn’t satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define dd double
using namespace std;

int main() {
	ios::sync_with_stdio(0);
	ll n; cin >> n;
	ll m = n / 2;
	m++;
	ll t = 1;
	cout << m << endl;
	for (ll i = 1; i <= m; i++) {
		cout << 1 << " " << i << endl;
	}
	for (ll i = 2; i <= n - m + 1; i++) {
		cout << i << " " << m << endl;
	}
}