#include <bits/stdc++.h>
using namespace std;
long long int lx[100000];
long long int rx[100000];
vector<pair<long long int, long long int>> leftx;
vector<pair<long long int, long long int>> rightx;
bool comp1(pair<long long int, long long int> lhs,
           pair<long long int, long long int> rhs) {
  return (lhs.second < rhs.second) ||
         (lhs.second == rhs.second && rx[lhs.first] < rx[rhs.first]);
}
bool comp2(pair<long long int, long long int> lhs,
           pair<long long int, long long int> rhs) {
  return (lhs.second < rhs.second) ||
         (lhs.second == rhs.second && lx[lhs.first] < lx[rhs.first]);
}
int main() {
  long long int n, xl, xr;
  scanf("%I64d%I64d%I64d", &n, &xl, &xr);
  for (int i = 0; i <= (int)n - 1; ++i) {
    long long int m, c;
    scanf("%I64d%I64d", &m, &c);
    lx[i] = xl * m + c;
    rx[i] = xr * m + c;
    leftx.push_back(make_pair(i, lx[i]));
    rightx.push_back(make_pair(i, rx[i]));
  }
  sort(leftx.begin(), leftx.end(), comp1);
  sort(rightx.begin(), rightx.end(), comp2);
  int f = 0;
  for (int i = 0; i <= (int)n - 1; ++i) {
    if (leftx[i].first != rightx[i].first) {
      f = 1;
      break;
    }
  }
  puts(f ? "Yes" : "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;
long long int lx[100000];
long long int rx[100000];
vector<pair<long long int, long long int>> leftx;
vector<pair<long long int, long long int>> rightx;
bool comp1(pair<long long int, long long int> lhs,
           pair<long long int, long long int> rhs) {
  return (lhs.second < rhs.second) ||
         (lhs.second == rhs.second && rx[lhs.first] < rx[rhs.first]);
}
bool comp2(pair<long long int, long long int> lhs,
           pair<long long int, long long int> rhs) {
  return (lhs.second < rhs.second) ||
         (lhs.second == rhs.second && lx[lhs.first] < lx[rhs.first]);
}
int main() {
  long long int n, xl, xr;
  scanf("%I64d%I64d%I64d", &n, &xl, &xr);
  for (int i = 0; i <= (int)n - 1; ++i) {
    long long int m, c;
    scanf("%I64d%I64d", &m, &c);
    lx[i] = xl * m + c;
    rx[i] = xr * m + c;
    leftx.push_back(make_pair(i, lx[i]));
    rightx.push_back(make_pair(i, rx[i]));
  }
  sort(leftx.begin(), leftx.end(), comp1);
  sort(rightx.begin(), rightx.end(), comp2);
  int f = 0;
  for (int i = 0; i <= (int)n - 1; ++i) {
    if (leftx[i].first != rightx[i].first) {
      f = 1;
      break;
    }
  }
  puts(f ? "Yes" : "No");
  return 0;
}

```