#include <bits/stdc++.h>
using namespace std;
pair<pair<long long int, long long int>, pair<int, int> > ranges[100001];
int n, x1, x2;
int main() {
  scanf("%d", &n);
  scanf("%d %d", &x1, &x2);
  for (int I = 0; I < n; I++) {
    int k, b;
    scanf("%d %d", &k, &b);
    long long int miny =
        ((long long int)k * (long long int)x1) + (long long int)b;
    long long int maxy =
        ((long long int)k * (long long int)x2) + (long long int)b;
    pair<long long int, long long int> p2 =
        make_pair(min(miny, maxy), max(miny, maxy));
    pair<int, int> p1 = make_pair(k, b);
    pair<pair<long long int, long long int>, pair<int, int> > p =
        make_pair(p2, p1);
    ranges[I] = p;
  }
  sort(ranges, ranges + n);
  pair<pair<long long int, long long int>, pair<int, int> > last = ranges[0];
  for (int I = 1; I < n; I++) {
    pair<pair<long long int, long long int>, pair<int, int> > rng = ranges[I];
    if (rng.first.second < last.first.second &&
        rng.first.first > last.first.first) {
      printf("YES");
      return 0;
    }
    if (rng.first.first == last.first.first &&
        rng.first.second == last.first.second) {
      printf("YES");
      return 0;
    }
    if ((last.second.first < 0 && rng.second.first > 0) ||
        (last.second.first > 0 && rng.second.first < 0)) {
      if (rng.first.first < last.first.second) {
        printf("YES");
        return 0;
      }
    }
    last = rng;
  }
  printf("NO");
}

### Prompt

Please create a solution in CPP 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;
pair<pair<long long int, long long int>, pair<int, int> > ranges[100001];
int n, x1, x2;
int main() {
  scanf("%d", &n);
  scanf("%d %d", &x1, &x2);
  for (int I = 0; I < n; I++) {
    int k, b;
    scanf("%d %d", &k, &b);
    long long int miny =
        ((long long int)k * (long long int)x1) + (long long int)b;
    long long int maxy =
        ((long long int)k * (long long int)x2) + (long long int)b;
    pair<long long int, long long int> p2 =
        make_pair(min(miny, maxy), max(miny, maxy));
    pair<int, int> p1 = make_pair(k, b);
    pair<pair<long long int, long long int>, pair<int, int> > p =
        make_pair(p2, p1);
    ranges[I] = p;
  }
  sort(ranges, ranges + n);
  pair<pair<long long int, long long int>, pair<int, int> > last = ranges[0];
  for (int I = 1; I < n; I++) {
    pair<pair<long long int, long long int>, pair<int, int> > rng = ranges[I];
    if (rng.first.second < last.first.second &&
        rng.first.first > last.first.first) {
      printf("YES");
      return 0;
    }
    if (rng.first.first == last.first.first &&
        rng.first.second == last.first.second) {
      printf("YES");
      return 0;
    }
    if ((last.second.first < 0 && rng.second.first > 0) ||
        (last.second.first > 0 && rng.second.first < 0)) {
      if (rng.first.first < last.first.second) {
        printf("YES");
        return 0;
      }
    }
    last = rng;
  }
  printf("NO");
}

```