#include <bits/stdc++.h>
using namespace std;
int n;
long long x1, x2;
long long k, b;
struct a {
  long long left, right, ID;
} arr[100000 + 16], arr1[100000 + 16];
;
bool cmp1(a m, a n) {
  if (m.left != n.left)
    return m.left < n.left;
  else
    return m.right < n.right;
}
bool cmp2(a m, a n) {
  if (m.right != n.right)
    return m.right < n.right;
  else
    return m.left < n.left;
}
int main(int argc, const char* argv[]) {
  bool f;
  while (scanf("%d", &n) == 1) {
    f = false;
    scanf("%I64d%I64d", &x1, &x2);
    for (int i = 1; i <= n; i++) {
      scanf("%I64d%I64d", &k, &b);
      arr[i].left = k * x1 + b;
      arr[i].right = k * x2 + b;
      arr[i].ID = i;
      arr1[i].left = k * x1 + b;
      arr1[i].right = k * x2 + b;
      arr1[i].ID = i;
    }
    sort(arr + 1, arr + n + 1, cmp1);
    sort(arr1 + 1, arr1 + n + 1, cmp2);
    for (int i = 1; i <= n; i++) {
      if (arr[i].ID != arr1[i].ID) f = true;
    }
    if (f)
      printf("Yes\n");
    else
      printf("No\n");
  }
  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;
int n;
long long x1, x2;
long long k, b;
struct a {
  long long left, right, ID;
} arr[100000 + 16], arr1[100000 + 16];
;
bool cmp1(a m, a n) {
  if (m.left != n.left)
    return m.left < n.left;
  else
    return m.right < n.right;
}
bool cmp2(a m, a n) {
  if (m.right != n.right)
    return m.right < n.right;
  else
    return m.left < n.left;
}
int main(int argc, const char* argv[]) {
  bool f;
  while (scanf("%d", &n) == 1) {
    f = false;
    scanf("%I64d%I64d", &x1, &x2);
    for (int i = 1; i <= n; i++) {
      scanf("%I64d%I64d", &k, &b);
      arr[i].left = k * x1 + b;
      arr[i].right = k * x2 + b;
      arr[i].ID = i;
      arr1[i].left = k * x1 + b;
      arr1[i].right = k * x2 + b;
      arr1[i].ID = i;
    }
    sort(arr + 1, arr + n + 1, cmp1);
    sort(arr1 + 1, arr1 + n + 1, cmp2);
    for (int i = 1; i <= n; i++) {
      if (arr[i].ID != arr1[i].ID) f = true;
    }
    if (f)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}

```