#include <bits/stdc++.h>
using namespace std;
struct PT {
  double x, y;
  PT() {}
  PT(double x, double y) : x(x), y(y) {}
  PT operator+(const PT &p) const { return PT(x + p.x, y + p.y); }
  PT operator-(const PT &p) const { return PT(x - p.x, y - p.y); }
  PT operator*(double c) const { return PT(x * c, y * c); }
  PT operator/(double c) const { return PT(x / c, y / c); }
};
double cross(PT p, PT q) { return p.x * q.y - p.y * q.x; }
bool operator<(const pair<PT, PT> &a, const pair<PT, PT> &b) {
  if (a.first.y < b.first.y) return true;
  if (a.first.y > b.first.y) return false;
  return a.second.y < b.second.y;
}
PT ComputeLineIntersection(pair<PT, PT> x, pair<PT, PT> y) {
  PT a = x.first;
  PT b = x.second - x.first;
  PT c = y.first - x.first;
  PT d = y.first - y.second;
  return a + b * cross(c, d) / cross(b, d);
}
double calcY(double k, double b, double x) { return k * x + b; }
int main() {
  int n;
  double x1, x2;
  cin >> n >> x1 >> x2;
  vector<pair<PT, PT> > lines;
  for (int i = 0; i < n; ++i) {
    double k, b;
    cin >> k >> b;
    PT ax1 = PT(x1, calcY(k, b, x1));
    PT ax2 = PT(x2, calcY(k, b, x2));
    lines.push_back(make_pair(ax1, ax2));
  }
  sort(lines.begin(), lines.end());
  double alpha = -1e100;
  for (auto l : lines) {
    if (l.second.y < alpha) {
      cout << "YES" << endl;
      return 0;
    }
    alpha = l.second.y;
  }
  cout << "NO" << endl;
}

### Prompt

Generate a cpp solution to 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;
struct PT {
  double x, y;
  PT() {}
  PT(double x, double y) : x(x), y(y) {}
  PT operator+(const PT &p) const { return PT(x + p.x, y + p.y); }
  PT operator-(const PT &p) const { return PT(x - p.x, y - p.y); }
  PT operator*(double c) const { return PT(x * c, y * c); }
  PT operator/(double c) const { return PT(x / c, y / c); }
};
double cross(PT p, PT q) { return p.x * q.y - p.y * q.x; }
bool operator<(const pair<PT, PT> &a, const pair<PT, PT> &b) {
  if (a.first.y < b.first.y) return true;
  if (a.first.y > b.first.y) return false;
  return a.second.y < b.second.y;
}
PT ComputeLineIntersection(pair<PT, PT> x, pair<PT, PT> y) {
  PT a = x.first;
  PT b = x.second - x.first;
  PT c = y.first - x.first;
  PT d = y.first - y.second;
  return a + b * cross(c, d) / cross(b, d);
}
double calcY(double k, double b, double x) { return k * x + b; }
int main() {
  int n;
  double x1, x2;
  cin >> n >> x1 >> x2;
  vector<pair<PT, PT> > lines;
  for (int i = 0; i < n; ++i) {
    double k, b;
    cin >> k >> b;
    PT ax1 = PT(x1, calcY(k, b, x1));
    PT ax2 = PT(x2, calcY(k, b, x2));
    lines.push_back(make_pair(ax1, ax2));
  }
  sort(lines.begin(), lines.end());
  double alpha = -1e100;
  for (auto l : lines) {
    if (l.second.y < alpha) {
      cout << "YES" << endl;
      return 0;
    }
    alpha = l.second.y;
  }
  cout << "NO" << endl;
}

```