#include <bits/stdc++.h>
using namespace std;
struct sort_pred {
  bool operator()(const std::pair<int, int> &left,
                  const std::pair<int, int> &right) {
    return left.second < right.second;
  }
};
long long k[100001];
long long b[100001];
int main() {
  int n;
  cin >> n;
  long long x1, x2;
  cin >> x1 >> x2;
  vector<pair<long long, long long> > y1;
  vector<pair<long long, long long> > y2;
  for (int i = 0; i < n; i++) {
    cin >> k[i] >> b[i];
    y1.push_back(make_pair(x1 * k[i] + b[i], i));
    y2.push_back(make_pair(x2 * k[i] + b[i], i));
  }
  sort(y1.begin(), y1.end());
  vector<pair<long long, long long> > y1th(n);
  vector<pair<long long, long long> > y2th(n);
  int th = 0;
  int i = 0;
  while (i < n) {
    int j = i;
    int cnt = 0;
    while (j < n) {
      cnt++;
      if (j == n - 1) break;
      if (y1[j].first != y1[j + 1].first) break;
      j++;
    }
    int k = i;
    while (k <= j) {
      y1th[y1[k].second] = make_pair(th, th + cnt - 1);
      k++;
    }
    th = th + cnt;
    i = j + 1;
  }
  sort(y2.begin(), y2.end());
  th = 0;
  i = 0;
  while (i < n) {
    int j = i;
    int cnt = 0;
    while (j < n) {
      cnt++;
      if (j == n - 1) break;
      if (y2[j].first != y2[j + 1].first) break;
      j++;
    }
    int k = i;
    while (k <= j) {
      y2th[y2[k].second] = make_pair(th, th + cnt - 1);
      k++;
    }
    th = th + cnt;
    i = j + 1;
  }
  for (int i = 0; i < n; i++) {
    if (y1th[i].second < y2th[i].first || y2th[i].second < y1th[i].first) {
      printf("YES");
      return 0;
    }
  }
  printf("NO");
  return 0;
}

### 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 sort_pred {
  bool operator()(const std::pair<int, int> &left,
                  const std::pair<int, int> &right) {
    return left.second < right.second;
  }
};
long long k[100001];
long long b[100001];
int main() {
  int n;
  cin >> n;
  long long x1, x2;
  cin >> x1 >> x2;
  vector<pair<long long, long long> > y1;
  vector<pair<long long, long long> > y2;
  for (int i = 0; i < n; i++) {
    cin >> k[i] >> b[i];
    y1.push_back(make_pair(x1 * k[i] + b[i], i));
    y2.push_back(make_pair(x2 * k[i] + b[i], i));
  }
  sort(y1.begin(), y1.end());
  vector<pair<long long, long long> > y1th(n);
  vector<pair<long long, long long> > y2th(n);
  int th = 0;
  int i = 0;
  while (i < n) {
    int j = i;
    int cnt = 0;
    while (j < n) {
      cnt++;
      if (j == n - 1) break;
      if (y1[j].first != y1[j + 1].first) break;
      j++;
    }
    int k = i;
    while (k <= j) {
      y1th[y1[k].second] = make_pair(th, th + cnt - 1);
      k++;
    }
    th = th + cnt;
    i = j + 1;
  }
  sort(y2.begin(), y2.end());
  th = 0;
  i = 0;
  while (i < n) {
    int j = i;
    int cnt = 0;
    while (j < n) {
      cnt++;
      if (j == n - 1) break;
      if (y2[j].first != y2[j + 1].first) break;
      j++;
    }
    int k = i;
    while (k <= j) {
      y2th[y2[k].second] = make_pair(th, th + cnt - 1);
      k++;
    }
    th = th + cnt;
    i = j + 1;
  }
  for (int i = 0; i < n; i++) {
    if (y1th[i].second < y2th[i].first || y2th[i].second < y1th[i].first) {
      printf("YES");
      return 0;
    }
  }
  printf("NO");
  return 0;
}

```