Codeforces Round #395(Div. 2)D. Timofey and rectangles

四色问题的内容是“任何一张地图只用四种颜色就能使具有共同边界的国家着上不同的颜色。”也就是说在不引起混淆的情况下一张地图只需四种颜色来标记就行。

From-百度百科

昨天遇到一个有关的问题先贴在这里，以后做到新的问题再做总结

Codeforces Round #395 (Div. 2) D. 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

代码

#include <iostream>
#include<cstdio>
#include<stdlib.h>
using namespace std;
int main()
{
int n,x1,y1,x2,y2;
scanf("%d",&n);
printf("YES\n");
for(int i=0;i<n;i++)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
printf("%d\n",(abs(x1)%2)+1+(abs(y1)%2)*2);
}
return 0;
}