#include <bits/stdc++.h>
using namespace std;
using int64 = long long int;
int main() {
  int n;
  int x1, x2;
  vector<pair<int64, int64>> a;
  cin >> n >> x1 >> x2;
  for (int i = 0; i < n; ++i) {
    int64 k, b;
    cin >> k >> b;
    a.push_back({k * x1 + b, k * x2 + b});
  }
  sort(a.begin(), a.end());
  int64 ymax = numeric_limits<int64>::min();
  bool can = false;
  for (auto &p : a) {
    if (p.second < ymax) can = true;
    ymax = max(ymax, p.second);
  }
  if (can)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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;
using int64 = long long int;
int main() {
  int n;
  int x1, x2;
  vector<pair<int64, int64>> a;
  cin >> n >> x1 >> x2;
  for (int i = 0; i < n; ++i) {
    int64 k, b;
    cin >> k >> b;
    a.push_back({k * x1 + b, k * x2 + b});
  }
  sort(a.begin(), a.end());
  int64 ymax = numeric_limits<int64>::min();
  bool can = false;
  for (auto &p : a) {
    if (p.second < ymax) can = true;
    ymax = max(ymax, p.second);
  }
  if (can)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

```