#include <bits/stdc++.h>
using namespace std;
int GCD(int a, int b) { return b ? GCD(b, a % b) : a; }
bool chk(string first, string second) {
  string t1 = first + second;
  string t2 = second + first;
  return t1 < t2;
}
struct sort_pred {
  bool operator()(const pair<int, int> &left, const pair<int, int> &right) {
    return left.second < right.second;
  }
};
long long POW(long long Base, long long Exp) {
  long long y, ret = 1;
  y = Base;
  while (Exp) {
    if (Exp & 1) ret = (ret * y) % 1000000007;
    y = (y * y) % 1000000007;
    Exp /= 2;
  }
  return ret % 1000000007;
}
vector<long long> A, B, C, Res, Mark;
string str, str1, s1, s2;
set<string> st;
vector<string> chlo, final;
vector<pair<long long, long long> > Rec;
int main() {
  long long n, i, j, tmp1, tmp2;
  cin >> n;
  Rec.resize(n);
  long long x1, x2;
  cin >> x1 >> x2;
  for (i = 0; i < n; i++) {
    cin >> tmp1 >> tmp2;
    Rec[i].first = tmp1 * x1 + tmp2;
    Rec[i].second = tmp1 * x2 + tmp2;
  }
  int flag = 0;
  sort(Rec.begin(), Rec.end());
  for (i = 1; i < n; i++) {
    if (Rec[i].second < Rec[i - 1].second) {
      flag = 1;
      break;
    }
  }
  if (!flag)
    printf("NO\n");
  else
    printf("YES\n");
  return 0;
}

### 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;
int GCD(int a, int b) { return b ? GCD(b, a % b) : a; }
bool chk(string first, string second) {
  string t1 = first + second;
  string t2 = second + first;
  return t1 < t2;
}
struct sort_pred {
  bool operator()(const pair<int, int> &left, const pair<int, int> &right) {
    return left.second < right.second;
  }
};
long long POW(long long Base, long long Exp) {
  long long y, ret = 1;
  y = Base;
  while (Exp) {
    if (Exp & 1) ret = (ret * y) % 1000000007;
    y = (y * y) % 1000000007;
    Exp /= 2;
  }
  return ret % 1000000007;
}
vector<long long> A, B, C, Res, Mark;
string str, str1, s1, s2;
set<string> st;
vector<string> chlo, final;
vector<pair<long long, long long> > Rec;
int main() {
  long long n, i, j, tmp1, tmp2;
  cin >> n;
  Rec.resize(n);
  long long x1, x2;
  cin >> x1 >> x2;
  for (i = 0; i < n; i++) {
    cin >> tmp1 >> tmp2;
    Rec[i].first = tmp1 * x1 + tmp2;
    Rec[i].second = tmp1 * x2 + tmp2;
  }
  int flag = 0;
  sort(Rec.begin(), Rec.end());
  for (i = 1; i < n; i++) {
    if (Rec[i].second < Rec[i - 1].second) {
      flag = 1;
      break;
    }
  }
  if (!flag)
    printf("NO\n");
  else
    printf("YES\n");
  return 0;
}

```