#include <bits/stdc++.h>
using namespace std;
long long line(int k, int b, int x) { return x * (long long)k + b; }
int main(int argc, char* argv[]) {
  int N;
  cin >> N;
  int b1, b2;
  cin >> b1 >> b2;
  map<long long, pair<long long, long long> > mapping;
  bool yes = false;
  for (int i = 0; i < N; i++) {
    int k, b;
    cin >> k >> b;
    long long y1 = line(k, b, b1);
    long long y2 = line(k, b, b2);
    if (mapping.empty()) {
      mapping[y1] = pair<long long, long long>(y2, y2);
    } else {
      map<long long, pair<long long, long long> >::iterator it;
      if (mapping.find(y1) != mapping.end()) {
        it = mapping.find(y1);
        if (y2 < it->second.first) it->second.first = y2;
        if (y2 > it->second.second) {
          it->second.second = y2;
          map<long long, pair<long long, long long> >::iterator it2 = it;
          it2++;
          if (it != mapping.end() && it2 != mapping.end() &&
              it2->second.first < it->second.second) {
            yes = true;
            break;
          }
        }
      } else {
        mapping[y1] = pair<long long, long long>(y2, y2);
        it = mapping.find(y1);
      }
      map<long long, pair<long long, long long> >::iterator itNext = it;
      itNext++;
      if (it != mapping.end() && itNext != mapping.end()) {
        if (itNext->second.first < it->second.second) {
          yes = true;
          break;
        }
      }
      itNext--;
      if (itNext != mapping.begin()) {
        itNext--;
        if (itNext->second.second > it->second.first) {
          yes = true;
          break;
        }
      }
    }
  }
  if (yes)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

### 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;
long long line(int k, int b, int x) { return x * (long long)k + b; }
int main(int argc, char* argv[]) {
  int N;
  cin >> N;
  int b1, b2;
  cin >> b1 >> b2;
  map<long long, pair<long long, long long> > mapping;
  bool yes = false;
  for (int i = 0; i < N; i++) {
    int k, b;
    cin >> k >> b;
    long long y1 = line(k, b, b1);
    long long y2 = line(k, b, b2);
    if (mapping.empty()) {
      mapping[y1] = pair<long long, long long>(y2, y2);
    } else {
      map<long long, pair<long long, long long> >::iterator it;
      if (mapping.find(y1) != mapping.end()) {
        it = mapping.find(y1);
        if (y2 < it->second.first) it->second.first = y2;
        if (y2 > it->second.second) {
          it->second.second = y2;
          map<long long, pair<long long, long long> >::iterator it2 = it;
          it2++;
          if (it != mapping.end() && it2 != mapping.end() &&
              it2->second.first < it->second.second) {
            yes = true;
            break;
          }
        }
      } else {
        mapping[y1] = pair<long long, long long>(y2, y2);
        it = mapping.find(y1);
      }
      map<long long, pair<long long, long long> >::iterator itNext = it;
      itNext++;
      if (it != mapping.end() && itNext != mapping.end()) {
        if (itNext->second.first < it->second.second) {
          yes = true;
          break;
        }
      }
      itNext--;
      if (itNext != mapping.begin()) {
        itNext--;
        if (itNext->second.second > it->second.first) {
          yes = true;
          break;
        }
      }
    }
  }
  if (yes)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

```