#include <bits/stdc++.h>
using namespace std;
double esp = 1e-12;
#pragma comment(linker, "/STACK:1024000000,1024000000")
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const long long int MOD = 1000000007;
vector<pair<long long int, long long int> > q[2];
int m[2][100001];
int main() {
  int n;
  scanf("%d", &n);
  q[0].clear();
  q[1].clear();
  int x1, x2, u, v;
  scanf("%d%d", &x1, &x2);
  for (int x = 1; x <= n; x++) {
    scanf("%d%d", &u, &v);
    q[0].push_back(
        make_pair((long long int)u * (long long int)x1 + v, (long long int)x));
    q[1].push_back(
        make_pair((long long int)u * (long long int)x2 + v, (long long int)x));
  }
  int flag = 0;
  sort(q[0].begin(), q[0].end());
  sort(q[1].begin(), q[1].end());
  for (int x = 0; x < n; x++) {
    m[0][q[0][x].second] = x;
    m[1][q[1][x].second] = x;
  }
  for (int x = 0; x < n; x++) {
    if (q[0][x].second != q[1][x].second) {
      if (q[1][m[1][q[0][x].second]].first == q[1][x].first) {
        swap(q[1][x], q[1][m[1][q[0][x].second]]);
        continue;
      }
      if (q[0][m[0][q[1][x].second]].first == q[0][x].first) {
        swap(q[0][x], q[0][m[0][q[1][x].second]]);
        continue;
      } else {
        flag = 1;
        break;
      }
    }
  }
  if (flag == 1)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

### 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;
double esp = 1e-12;
#pragma comment(linker, "/STACK:1024000000,1024000000")
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const long long int MOD = 1000000007;
vector<pair<long long int, long long int> > q[2];
int m[2][100001];
int main() {
  int n;
  scanf("%d", &n);
  q[0].clear();
  q[1].clear();
  int x1, x2, u, v;
  scanf("%d%d", &x1, &x2);
  for (int x = 1; x <= n; x++) {
    scanf("%d%d", &u, &v);
    q[0].push_back(
        make_pair((long long int)u * (long long int)x1 + v, (long long int)x));
    q[1].push_back(
        make_pair((long long int)u * (long long int)x2 + v, (long long int)x));
  }
  int flag = 0;
  sort(q[0].begin(), q[0].end());
  sort(q[1].begin(), q[1].end());
  for (int x = 0; x < n; x++) {
    m[0][q[0][x].second] = x;
    m[1][q[1][x].second] = x;
  }
  for (int x = 0; x < n; x++) {
    if (q[0][x].second != q[1][x].second) {
      if (q[1][m[1][q[0][x].second]].first == q[1][x].first) {
        swap(q[1][x], q[1][m[1][q[0][x].second]]);
        continue;
      }
      if (q[0][m[0][q[1][x].second]].first == q[0][x].first) {
        swap(q[0][x], q[0][m[0][q[1][x].second]]);
        continue;
      } else {
        flag = 1;
        break;
      }
    }
  }
  if (flag == 1)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

```