#include <bits/stdc++.h>
using namespace std;
struct Compare {
  bool operator()(pair<long long, long long> &l,
                  pair<long long, long long> &r) {
    if (min(l.first, l.second) == min(r.first, r.second)) {
      return max(l.first, l.second) < max(r.first, r.second);
    } else {
      return min(l.first, l.second) < min(r.first, r.second);
    }
  }
};
signed main() {
  ios::sync_with_stdio(0);
  cin.sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  long long x1, x2;
  cin >> x1 >> x2;
  long long k[n + 1], b[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> k[i] >> b[i];
  }
  if (false)
    cerr << ("k[1] * x1 + b[1]") << " is " << (k[1] * x1 + b[1]) << '\n';
  vector<pair<long long, long long> > all;
  for (long long i = 1; i <= n; i++) {
    if (false) cerr << "on " << k[i] << " " << b[i] << '\n';
    long long f = (k[i] * x1) + b[i];
    long long s = (k[i] * x2) + b[i];
    if (false) cerr << f << "," << s << '\n';
    all.emplace_back(pair<long long, long long>(f, s));
  }
  sort(all.begin(), all.end(), Compare());
  for (long long i = 0; i < all.size() - 1; i++) {
    long long imi = min(all[i].first, all[i].second);
    long long ima = max(all[i].first, all[i].second);
    long long nmi = min(all[i + 1].first, all[i + 1].second);
    long long nma = max(all[i + 1].first, all[i + 1].second);
    if ((all[i].first == all[i + 1].first) or
        (all[i].second == all[i + 1].second)) {
      if (false) cerr << "one is same" << '\n';
      continue;
    }
    if (not((all[i].first < all[i + 1].first) ^
            (all[i].second < all[i + 1].second))) {
      if (false) cerr << "stopped" << '\n';
      continue;
    }
    if (imi < nma and ima > nmi) {
      cout << "YES" << '\n';
      return 0;
    }
  }
  cout << "NO" << '\n';
}

### Prompt

In CPP, your task is to solve 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 Compare {
  bool operator()(pair<long long, long long> &l,
                  pair<long long, long long> &r) {
    if (min(l.first, l.second) == min(r.first, r.second)) {
      return max(l.first, l.second) < max(r.first, r.second);
    } else {
      return min(l.first, l.second) < min(r.first, r.second);
    }
  }
};
signed main() {
  ios::sync_with_stdio(0);
  cin.sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  long long x1, x2;
  cin >> x1 >> x2;
  long long k[n + 1], b[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> k[i] >> b[i];
  }
  if (false)
    cerr << ("k[1] * x1 + b[1]") << " is " << (k[1] * x1 + b[1]) << '\n';
  vector<pair<long long, long long> > all;
  for (long long i = 1; i <= n; i++) {
    if (false) cerr << "on " << k[i] << " " << b[i] << '\n';
    long long f = (k[i] * x1) + b[i];
    long long s = (k[i] * x2) + b[i];
    if (false) cerr << f << "," << s << '\n';
    all.emplace_back(pair<long long, long long>(f, s));
  }
  sort(all.begin(), all.end(), Compare());
  for (long long i = 0; i < all.size() - 1; i++) {
    long long imi = min(all[i].first, all[i].second);
    long long ima = max(all[i].first, all[i].second);
    long long nmi = min(all[i + 1].first, all[i + 1].second);
    long long nma = max(all[i + 1].first, all[i + 1].second);
    if ((all[i].first == all[i + 1].first) or
        (all[i].second == all[i + 1].second)) {
      if (false) cerr << "one is same" << '\n';
      continue;
    }
    if (not((all[i].first < all[i + 1].first) ^
            (all[i].second < all[i + 1].second))) {
      if (false) cerr << "stopped" << '\n';
      continue;
    }
    if (imi < nma and ima > nmi) {
      cout << "YES" << '\n';
      return 0;
    }
  }
  cout << "NO" << '\n';
}

```