## USACO 2015 February Contest, Gold

## Problem 3. Fencing the Herd

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Farmer John needs your help deciding where to build a fence in the shape of a straight line to help restrict the movement of his cows. He has considered several possible fence locations and needs your help to determine which of these are usable, where a fence is considered usable if all of the cows are on the same side of the fence. A fence is not usable if there is a cow that lies directly on it. FJ will be asking you a number of queries regarding possible fence locations; a query should be answered "YES" if it corresponds to a usable fence location, "NO" otherwise.

Additionally, FJ may occasionally bring new cows into the herd. When a new cow joins the herd, all fence queries from that point onward will require her to be on the same side of a fence as the rest of the herd for the fence to be usable.

#### INPUT FORMAT: (file fencing.in)

The first line of input contains N (1 <= N <= 100,000) and Q (1 <= Q <= 100,000) separated by a space. These give the number of cows initially in the herd and the number of queries, respectively.

The following N lines describe the initial state of the herd. Each line will contain two space separated integers x and y giving the position of a cow.

The remaining Q lines contain queries either adding a new cow to the herd or testing a fence for usability. A line of the form "1 x y" means that a new cow has been added to the herd at position (x, y). A line of the form "2 A B C" indicates that FJ would like to test a fence described by the line Ax + By = C.

All cow positions will be unique over the whole data set and will satisfy (-10^9 <= x, y <= 10^9). Additionally the fence queries will satisfy -10^9 <= A, B <= 10^9 and -10^18 <= C <= 10^18. No fence query will have A = B = 0.

#### OUTPUT FORMAT: (file fencing.out)

For each fence query, output "YES" if the fence is usable. Otherwise output "NO".

#### SAMPLE INPUT:

3 4 0 0 0 1 1 0 2 2 2 3 1 1 1 2 2 2 3 2 0 1 1

#### SAMPLE OUTPUT:

YES NO NO

The line 2x + 2y = 3 leaves the initial 3 cows on the same side. However the cow (1, 1) is on the other side of this fence making it no longer usable after she joins the herd. The line Y = 1 cannot be used because the cows (0, 1) and (1, 1) lie directly on it.

*Warning: The I/O for this problem is fairly large. C++ users may consider using scanf or the line "ios_base::sync_with_stdio(false)" to read input faster. Java users should avoid using java.util.Scanner. Do not flush output (e.g. using std::endl) after each query.*

[Problem credits: Richard Peng, 2015]