CodeForces - 764D Timofey and rectangles （四色定理+找規律）

題目鏈接：http://codeforces.com/problemset/problem/764/D點擊打開鏈接

D. Timofey and rectangles

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

因爲四色定理因此一定能構造出來

然後因爲是奇數長度因此相鄰兩個方塊橫座標或縱座標必有一個奇偶不同

由此可構造

#include <bits/stdc++.h>
using namespace std;
struct xjy
{
    int x1,x2;
    int y1,y2;
};
int main()
{
    int n;
    cin >> n ;
    cout << "YES" <<endl;
    for(int i=0;i<n;i++)
    {
        xjy mid;
        cin >> mid.x1 >> mid.y1 >> mid.x2 >> mid.y2;
        cout << ((mid.x1%2+mid.y1%2*2)+4)%4+1 <<endl;
    }
}