#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> p[100005];
int main() {
  long long n, x1, x2;
  scanf("%lld", &n);
  scanf("%lld%lld", &x1, &x2);
  int a, b;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a, &b);
    p[i].first = a * x1 + b, p[i].second = a * x2 + b;
  }
  sort(p, p + n);
  int temp = 0;
  for (int i = 0; i < n - 1; i++) {
    if (p[i].second > p[i + 1].second) {
      temp = 1;
      break;
    }
  }
  if (temp)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = ki·x + bi. 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' = ki * x' + bi, that is, point (x', y') belongs to the line number i; 
  * y' = kj * x' + bj, 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 ki, bi ( - 1 000 000 ≤ ki, bi ≤ 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 ki ≠ kj, or bi ≠ bj.

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

Note

In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> p[100005];
int main() {
  long long n, x1, x2;
  scanf("%lld", &n);
  scanf("%lld%lld", &x1, &x2);
  int a, b;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a, &b);
    p[i].first = a * x1 + b, p[i].second = a * x2 + b;
  }
  sort(p, p + n);
  int temp = 0;
  for (int i = 0; i < n - 1; i++) {
    if (p[i].second > p[i + 1].second) {
      temp = 1;
      break;
    }
  }
  if (temp)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}

```