B. Timofey and rectangles（四色定理，坐标奇偶错位）

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

80 0 5 32 -1 5 0-3 -4 2 -1-1 -1 2 0-3 0 0 55 2 10 37 -3 10 24 -2 7 -1

output

YES12232241

思路：因为是奇数，所以满足四色定理（可以自行百度）。也就是说把平面上不相交的边长是奇数的矩形染成不同的颜色，不超过四种。

然后就是，因为边长是奇数，（接下来是重点）：假设一个矩形的左下角坐标组合奇偶性为（奇A，奇A），那么右上角一定是（偶A，偶A）；

假设此时又有一矩形左下坐标是（奇B，奇B），那么他的右上角肯定是（偶B，偶B），由于（奇A，奇A），所以肯定不会挨着。因为恰好

有四种，（奇，偶）啥啥的，所以刚好是四种颜色

代码：

int main()  {      int a,b,c,d;      cin>>n;      cout<<"YES"<<endl;       while(n--)      {          cin >> a>> b>>c>>d;          if(a%2&&b%2)              cout << 1<<endl;          else if(a%2==0&&b%2)              cout << 2<<endl;          else if(a%2&&b%2==0)              cout << 3<<endl;          else              cout << 4<<endl;      }      return 0;     }