CodeForces - 764D Timofey and rectangles

One of Timofey's birthday presents is a colourbook in a shape of an infinite plane. On the plane n rectangles with sides parallel to coordinate axes are situated. All sides of the rectangles have odd length. Rectangles cannot intersect, but they can touch each other.

Help Timofey to color his rectangles in 4 different colors in such a way that every two rectangles touching each other by side would have different color, or determine that it is impossible.

Two rectangles intersect if their intersection has positive area. Two rectangles touch by sides if there is a pair of sides such that their intersection has non-zero length

The picture corresponds to the first example

Input

The first line contains single integer n (1 ≤ n ≤ 5·105) — the number of rectangles.

n lines follow. The i-th of these lines contains four integers x1, y1, x2 and y2 ( - 109 ≤ x1 < x2 ≤ 109,  - 109 ≤ y1 < y2 ≤ 109), that means that points (x1, y1) and (x2, y2) are the coordinates of two opposite corners of the i-th rectangle.

It is guaranteed, that all sides of the rectangles have odd lengths and rectangles don't intersect each other.

Output

Print "NO" in the only line if it is impossible to color the rectangles in 4 different colors in such a way that every two rectangles touching each other by side would have different color.

Otherwise, print "YES" in the first line. Then print n lines, in the i-th of them print single integer ci (1 ≤ ci ≤ 4) — the color of i-th rectangle.

Example

Input

8
0 0 5 3
2 -1 5 0
-3 -4 2 -1
-1 -1 2 0
-3 0 0 5
5 2 10 3
7 -3 10 2
4 -2 7 -1

Output

YES
1
2
2
3
2
2
4
1

题意：给定n个矩形，边长为奇数，相邻矩形的颜色不能为一样的

题解：四色定理：https://baike.baidu.com/item/%E5%9B%9B%E8%89%B2%E5%AE%9A%E7%90%86/805159?fr=aladdin

因为边长为奇数，我们可以确定右上点，来判断它的奇偶性，相邻矩形该点的x，y奇偶性一定不完全相同

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n,x,y;
    cin>>n;
    cout<<"YES"<<endl;
    while(n--)
    {
    	cin>>x>>y>>x>>y;
    	x=abs(x);
    	y=abs(y);
    	if(x&1&&y&1)cout<<"1"<<endl;
    	if(x&1&&!(y&1)) cout<<"2"<<endl;
    	if(!(x&1)&&y&1) cout<<"3"<<endl;
    	if(!(x&1)&&!(y&1)) cout<<"4"<<endl;
	}
    return 0;
}