#include <bits/stdc++.h>
using namespace std;
bool isZero(double v) { return abs(v) < 1e-10; }
class LineY {
 public:
  double yLeft, yRight;
  LineY(double yLeft, double yRight) {
    this->yLeft = yLeft;
    this->yRight = yRight;
  }
  bool operator<(const LineY &l2) const {
    if (!isZero(yLeft - l2.yLeft)) return yLeft > l2.yLeft;
    if (!isZero(yRight - l2.yRight)) return yRight > l2.yRight;
    return false;
  }
};
vector<LineY> lines;
int n;
bool intersectionFound() {
  sort(lines.begin(), lines.end());
  double lowestYRight = 1e20;
  for (int i = 0; i < lines.size(); i++) {
    double yRight = lines[i].yRight;
    if (yRight > lowestYRight + 1e-10) return true;
    lowestYRight = min(lowestYRight, yRight);
  }
  return false;
}
int main() {
  double xLeft, xRight;
  scanf("%d%lf%lf", &n, &xLeft, &xRight);
  for (int i = 0; i < n; i++) {
    double k, b;
    scanf("%lf%lf", &k, &b);
    double yLeft = k * xLeft + b;
    double yRight = k * xRight + b;
    lines.push_back(LineY(yLeft, yRight));
  }
  bool res = intersectionFound();
  if (res)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

### Prompt

Your challenge is to write 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;
bool isZero(double v) { return abs(v) < 1e-10; }
class LineY {
 public:
  double yLeft, yRight;
  LineY(double yLeft, double yRight) {
    this->yLeft = yLeft;
    this->yRight = yRight;
  }
  bool operator<(const LineY &l2) const {
    if (!isZero(yLeft - l2.yLeft)) return yLeft > l2.yLeft;
    if (!isZero(yRight - l2.yRight)) return yRight > l2.yRight;
    return false;
  }
};
vector<LineY> lines;
int n;
bool intersectionFound() {
  sort(lines.begin(), lines.end());
  double lowestYRight = 1e20;
  for (int i = 0; i < lines.size(); i++) {
    double yRight = lines[i].yRight;
    if (yRight > lowestYRight + 1e-10) return true;
    lowestYRight = min(lowestYRight, yRight);
  }
  return false;
}
int main() {
  double xLeft, xRight;
  scanf("%d%lf%lf", &n, &xLeft, &xRight);
  for (int i = 0; i < n; i++) {
    double k, b;
    scanf("%lf%lf", &k, &b);
    double yLeft = k * xLeft + b;
    double yRight = k * xRight + b;
    lines.push_back(LineY(yLeft, yRight));
  }
  bool res = intersectionFound();
  if (res)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

```