#include <bits/stdc++.h>
using namespace std;
double A2[100001][3], A1[100001][3], B[100001][3];
void merge_sort2(long long int l, long long int r) {
  long long int m = (l + r) >> 1, i, j, k;
  if (l == r) return;
  merge_sort2(l, m);
  merge_sort2(m + 1, r);
  for (i = l, j = m + 1, k = l; i <= m || j <= r;)
    if (j > r || (i <= m && A2[i][0] < A2[j][0])) {
      B[k][0] = A2[i][0];
      B[k][2] = A2[i][2];
      k++;
      i++;
    } else {
      B[k][0] = A2[j][0];
      B[k][2] = A2[j][2];
      k++;
      j++;
    }
  for (k = l; k <= r; k++) {
    A2[k][0] = B[k][0];
    A2[k][2] = B[k][2];
  }
}
void merge_sort1(long long int l, long long int r) {
  long long int m = (l + r) >> 1, i, j, k;
  if (l == r) return;
  merge_sort1(l, m);
  merge_sort1(m + 1, r);
  for (i = l, j = m + 1, k = l; i <= m || j <= r;)
    if (j > r || (i <= m && A1[i][0] < A1[j][0])) {
      B[k][0] = A1[i][0];
      B[k][2] = A1[i][2];
      k++;
      i++;
    } else {
      B[k][0] = A1[j][0];
      B[k][2] = A1[j][2];
      k++;
      j++;
    }
  for (k = l; k <= r; k++) {
    A1[k][0] = B[k][0];
    A1[k][2] = B[k][2];
  }
}
int main() {
  int i, n, k, b;
  double x1, x2;
  cin >> n >> x1 >> x2;
  x1 += 0.0000001;
  x2 -= 0.0000001;
  for (i = 1; i <= n; i++) {
    scanf("%i %i", &k, &b);
    A1[i][0] = k * x1 + b;
    A1[i][2] = i;
    A2[i][0] = k * x2 + b;
    A2[i][2] = i;
  }
  merge_sort1(1, n);
  merge_sort2(1, n);
  for (i = 1; i <= n; i++) {
    if (A1[i][2] != A2[i][2]) {
      printf("YES");
      return 0;
    }
  }
  printf("NO");
}

### Prompt

Generate 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;
double A2[100001][3], A1[100001][3], B[100001][3];
void merge_sort2(long long int l, long long int r) {
  long long int m = (l + r) >> 1, i, j, k;
  if (l == r) return;
  merge_sort2(l, m);
  merge_sort2(m + 1, r);
  for (i = l, j = m + 1, k = l; i <= m || j <= r;)
    if (j > r || (i <= m && A2[i][0] < A2[j][0])) {
      B[k][0] = A2[i][0];
      B[k][2] = A2[i][2];
      k++;
      i++;
    } else {
      B[k][0] = A2[j][0];
      B[k][2] = A2[j][2];
      k++;
      j++;
    }
  for (k = l; k <= r; k++) {
    A2[k][0] = B[k][0];
    A2[k][2] = B[k][2];
  }
}
void merge_sort1(long long int l, long long int r) {
  long long int m = (l + r) >> 1, i, j, k;
  if (l == r) return;
  merge_sort1(l, m);
  merge_sort1(m + 1, r);
  for (i = l, j = m + 1, k = l; i <= m || j <= r;)
    if (j > r || (i <= m && A1[i][0] < A1[j][0])) {
      B[k][0] = A1[i][0];
      B[k][2] = A1[i][2];
      k++;
      i++;
    } else {
      B[k][0] = A1[j][0];
      B[k][2] = A1[j][2];
      k++;
      j++;
    }
  for (k = l; k <= r; k++) {
    A1[k][0] = B[k][0];
    A1[k][2] = B[k][2];
  }
}
int main() {
  int i, n, k, b;
  double x1, x2;
  cin >> n >> x1 >> x2;
  x1 += 0.0000001;
  x2 -= 0.0000001;
  for (i = 1; i <= n; i++) {
    scanf("%i %i", &k, &b);
    A1[i][0] = k * x1 + b;
    A1[i][2] = i;
    A2[i][0] = k * x2 + b;
    A2[i][2] = i;
  }
  merge_sort1(1, n);
  merge_sort2(1, n);
  for (i = 1; i <= n; i++) {
    if (A1[i][2] != A2[i][2]) {
      printf("YES");
      return 0;
    }
  }
  printf("NO");
}

```