#include <bits/stdc++.h>
using namespace std;
template <class T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
T my_pow(T n, T p) {
  if (p == 0) return 1;
  T x = my_pow(n, p / 2);
  x = (x * x);
  if (p & 1) x = (x * n);
  return x;
}
template <class T>
T big_mod(T n, T p, T m) {
  if (p == 0) return (T)1;
  T x = big_mod(n, p / 2, m);
  x = (x * x) % m;
  if (p & 1) x = (x * n) % m;
  return x;
}
template <class T>
inline T Mod(T n, T m) {
  return (n % m + m) % m;
}
template <class T>
T extract(string s, T ret) {
  stringstream ss(s);
  ss >> ret;
  return ret;
}
template <class T>
string itos(T n) {
  ostringstream ost;
  ost << n;
  ost.flush();
  return ost.str();
}
long long stoi(string s) {
  long long r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
double toDouble(string s) {
  double r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
struct info {
  long long x, y;
} arr[100005];
bool com(info a, info b) { return ((a.x < b.x) or (a.x == b.x and a.y < b.y)); }
int main() {
  long long n, x1, y1;
  cin >> n >> x1 >> y1;
  long long a, b;
  for (__typeof(n) i = 0; i < (n); i++) {
    cin >> a >> b;
    arr[i].x = a * x1 + b;
    arr[i].y = a * y1 + b;
  }
  sort(arr, arr + n, com);
  long long f = 0;
  long long mx = -100000000000000LL;
  for (int i = 0; i < n - 1; i++) {
    if (arr[i].x < arr[i + 1].x and arr[i + 1].y < arr[i].y) f = 1;
  }
  if (f)
    cout << "yes";
  else
    cout << "no";
  return 0;
}

### Prompt

Your task is to create 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;
template <class T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
T my_pow(T n, T p) {
  if (p == 0) return 1;
  T x = my_pow(n, p / 2);
  x = (x * x);
  if (p & 1) x = (x * n);
  return x;
}
template <class T>
T big_mod(T n, T p, T m) {
  if (p == 0) return (T)1;
  T x = big_mod(n, p / 2, m);
  x = (x * x) % m;
  if (p & 1) x = (x * n) % m;
  return x;
}
template <class T>
inline T Mod(T n, T m) {
  return (n % m + m) % m;
}
template <class T>
T extract(string s, T ret) {
  stringstream ss(s);
  ss >> ret;
  return ret;
}
template <class T>
string itos(T n) {
  ostringstream ost;
  ost << n;
  ost.flush();
  return ost.str();
}
long long stoi(string s) {
  long long r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
double toDouble(string s) {
  double r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
struct info {
  long long x, y;
} arr[100005];
bool com(info a, info b) { return ((a.x < b.x) or (a.x == b.x and a.y < b.y)); }
int main() {
  long long n, x1, y1;
  cin >> n >> x1 >> y1;
  long long a, b;
  for (__typeof(n) i = 0; i < (n); i++) {
    cin >> a >> b;
    arr[i].x = a * x1 + b;
    arr[i].y = a * y1 + b;
  }
  sort(arr, arr + n, com);
  long long f = 0;
  long long mx = -100000000000000LL;
  for (int i = 0; i < n - 1; i++) {
    if (arr[i].x < arr[i + 1].x and arr[i + 1].y < arr[i].y) f = 1;
  }
  if (f)
    cout << "yes";
  else
    cout << "no";
  return 0;
}

```