2020 BIT冬训-C++STL K - Anton and Lines CodeForces - 593B

Problem Description

The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces.

In the task he was given a set of n lines defined by the equations y = k_i·x + b_i. It was necessary to determine whether there is at least

one point of intersection of two of these lines, that lays strictly inside the strip between x₁ < x₂. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:

• y' = k_i * x' + b_i, that is, point (x', y') belongs to the line number i;
• y' = k_j * x' + b_j, that is, point (x', y') belongs to the line number j;
• x₁ < x' < x₂, that is, point (x', y') lies inside the strip bounded by x₁ < x₂.

You can't leave Anton in trouble, can you? Write a program that solves the given task.

 

Input

The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton.

The second line contains integers x₁ and x₂ ( - 1 000 000 ≤ x₁ < x₂ ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.

The following n lines contain integers k_i, b_i ( - 1 000 000 ≤ k_i, b_i ≤ 1 000 000) — the descriptions of the lines.

It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either k_i ≠ k_j, or b_i ≠ b_j.

 

Output

Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).

 

Examples

Input

4
1 2
1 2
1 0
0 1
0 2

Output

NO

Input

2
1 3
1 0
-1 3

Output

YES

Input

2
1 3
1 0
0 2

Output

YES

Input

2
1 3
1 0
0 3

Output

NO

这题主要是要开long long，int的话数据会爆导致错误nnd。
还有就是冷静分析区间内相交的性质。（零点存在性定理）

#include<stdio.h>
#include<algorithm>
using namespace std;
pair<__int64,__int64> line[100005];
__int64 n,x1,x2,k,b;
int main(){
    scanf("%I64d%I64d%I64d",&n,&x1,&x2);
    for(int i =0;i<n;i++){
        scanf("%I64d%I64d",&k,&b);
        line[i]={k*x1+b,k*x2+b};
    }
    sort(line,line+n);
    for(int i=1;i<n;i++){
        if(line[i-1].second>line[i].second){
            printf("YES");
            return 0;
        }
    }
    printf("NO");
    return 0;
}