593_B. Anton and Lines_8830
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>
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long i, k, b, n, x1, x2, prev1, prev2;
  cin >> n;
  cin >> x1 >> x2;
  vector<pair<long long, long long> > y(n);
  for (i = 0; i < n; i++) {
    cin >> k >> b;
    y[i] = make_pair(k * x1 + b, k * x2 + b);
  }
  sort(y.begin(), y.end());
  prev1 = y[0].first;
  prev2 = y[0].second;
  for (i = 1; i < n; i++) {
    if (y[i].second < prev2) {
      cout << "YES";
      return 0;
    }
    prev2 = max(prev2, y[i].second);
  }
  cout << "NO";
  return 0;
}
