#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
vector<pair<pair<double, double>, int> > sol;
vector<pair<double, double> > v;
int main() {
  long long n;
  double x1, x2;
  double k, b;
  cin >> n;
  cin >> x1 >> x2;
  for (int i = 0; i < n; ++i) {
    scanf("%lf %lf", &k, &b);
    double tmp1 = k * x1 + b, tmp2 = k * x2 + b;
    v.push_back(make_pair(k, b));
    sol.push_back(make_pair(make_pair(min(tmp1, tmp2), max(tmp1, tmp2)), i));
  }
  sort(sol.begin(), sol.end());
  bool ans = false;
  for (int i = 0; i < sol.size() && !ans; ++i) {
    double e = sol[i].first.second;
    int ind = sol[i].second;
    ++i;
    while (i < sol.size() && sol[i].first.first <= e) {
      long long ind2 = sol[i].second;
      double k1 = v[ind].first, b1 = v[ind].second, k2 = v[ind2].first,
             b2 = v[ind2].second;
      if (k1 != k2) {
        double x = (b1 - b2) / (k2 - k1);
        if (x > x1 && x < x2) ans = true;
      }
      ind = ind2;
      e = max(e, sol[i].first.second);
      ++i;
    }
    --i;
  }
  if (ans)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

### Prompt

Please formulate 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;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
vector<pair<pair<double, double>, int> > sol;
vector<pair<double, double> > v;
int main() {
  long long n;
  double x1, x2;
  double k, b;
  cin >> n;
  cin >> x1 >> x2;
  for (int i = 0; i < n; ++i) {
    scanf("%lf %lf", &k, &b);
    double tmp1 = k * x1 + b, tmp2 = k * x2 + b;
    v.push_back(make_pair(k, b));
    sol.push_back(make_pair(make_pair(min(tmp1, tmp2), max(tmp1, tmp2)), i));
  }
  sort(sol.begin(), sol.end());
  bool ans = false;
  for (int i = 0; i < sol.size() && !ans; ++i) {
    double e = sol[i].first.second;
    int ind = sol[i].second;
    ++i;
    while (i < sol.size() && sol[i].first.first <= e) {
      long long ind2 = sol[i].second;
      double k1 = v[ind].first, b1 = v[ind].second, k2 = v[ind2].first,
             b2 = v[ind2].second;
      if (k1 != k2) {
        double x = (b1 - b2) / (k2 - k1);
        if (x > x1 && x < x2) ans = true;
      }
      ind = ind2;
      e = max(e, sol[i].first.second);
      ++i;
    }
    --i;
  }
  if (ans)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

```