#include <bits/stdc++.h>
using namespace std;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const double PI = 3.14159265358979323846;
const double eps = (1e-7);
enum { LESS, EQUAL, GREATER };
int dcmp(long double x, long double y) {
  if (fabs(x - y) <= eps) return EQUAL;
  return x > y ? GREATER : LESS;
}
const int MAX_N = 2e5 + 5;
long long t[4 * MAX_N];
void update(int i, long long v, int p = 1, int l = 0, int r = MAX_N - 1) {
  if (l == r)
    t[p] = max(t[p], v);
  else {
    int mid = (l + r) / 2;
    if (i <= mid)
      update(i, v, 2 * p, l, mid);
    else
      update(i, v, 2 * p + 1, mid + 1, r);
    t[p] = max(t[2 * p], t[2 * p + 1]);
  }
}
long long get(int stq, int enq, int p = 1, int l = 0, int r = MAX_N - 1) {
  if (l >= stq && r <= enq) return t[p];
  if (l > enq || r < stq) return LLONG_MIN;
  int mid = (l + r) / 2;
  return max(get(stq, enq, 2 * p, l, mid),
             get(stq, enq, 2 * p + 1, mid + 1, r));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, x1, x2;
  cin >> n >> x1 >> x2;
  vector<pair<int, int> > line(n);
  vector<pair<long long, long long> > v(n);
  for (int i = 0; i < n; ++i) {
    cin >> line[i].first >> line[i].second;
    v[i].first = 1LL * line[i].first * (x1) + line[i].second;
    v[i].second = 1LL * line[i].first * (x2) + line[i].second;
  }
  sort(v.begin(), v.end());
  long long mx = LLONG_MIN;
  for (int i = 0; i < n;) {
    long long cur = v[i].first;
    int st = i;
    while (i < n && cur == v[i].first) {
      if (v[i].second < mx) {
        cout << "YES\n";
        return 0;
      }
      ++i;
    }
    while (st < i) mx = max(mx, v[st++].second);
  }
  cout << "NO\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const double PI = 3.14159265358979323846;
const double eps = (1e-7);
enum { LESS, EQUAL, GREATER };
int dcmp(long double x, long double y) {
  if (fabs(x - y) <= eps) return EQUAL;
  return x > y ? GREATER : LESS;
}
const int MAX_N = 2e5 + 5;
long long t[4 * MAX_N];
void update(int i, long long v, int p = 1, int l = 0, int r = MAX_N - 1) {
  if (l == r)
    t[p] = max(t[p], v);
  else {
    int mid = (l + r) / 2;
    if (i <= mid)
      update(i, v, 2 * p, l, mid);
    else
      update(i, v, 2 * p + 1, mid + 1, r);
    t[p] = max(t[2 * p], t[2 * p + 1]);
  }
}
long long get(int stq, int enq, int p = 1, int l = 0, int r = MAX_N - 1) {
  if (l >= stq && r <= enq) return t[p];
  if (l > enq || r < stq) return LLONG_MIN;
  int mid = (l + r) / 2;
  return max(get(stq, enq, 2 * p, l, mid),
             get(stq, enq, 2 * p + 1, mid + 1, r));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, x1, x2;
  cin >> n >> x1 >> x2;
  vector<pair<int, int> > line(n);
  vector<pair<long long, long long> > v(n);
  for (int i = 0; i < n; ++i) {
    cin >> line[i].first >> line[i].second;
    v[i].first = 1LL * line[i].first * (x1) + line[i].second;
    v[i].second = 1LL * line[i].first * (x2) + line[i].second;
  }
  sort(v.begin(), v.end());
  long long mx = LLONG_MIN;
  for (int i = 0; i < n;) {
    long long cur = v[i].first;
    int st = i;
    while (i < n && cur == v[i].first) {
      if (v[i].second < mx) {
        cout << "YES\n";
        return 0;
      }
      ++i;
    }
    while (st < i) mx = max(mx, v[st++].second);
  }
  cout << "NO\n";
}

```