#include <bits/stdc++.h>
using namespace std;
int n, x1, x2, k, b, aa;
long long as, q[100009], w[100009], e[100009], r[100009], t[100009];
void ma0(int x, long long a) {
  for (; x <= n; x += x & -x) {
    if (a > r[x]) r[x] = a;
  }
}
long long ma1(int x) {
  long long a = -10000000000000;
  for (; x > 0; x -= x & -x)
    if (r[x] > a) a = r[x];
  return a;
}
void mi0(int x, long long a) {
  for (; x <= n; x += x & -x) {
    if (a < t[x]) t[x] = a;
  }
}
long long mi1(int x) {
  long long a = 10000000000000;
  for (; x > 0; x -= x & -x)
    if (t[x] < a) a = t[x];
  return a;
}
int main() {
  scanf("%d%d%d", &n, &x1, &x2);
  for (int i = 1; i <= n; i++) {
    t[i] = 10000000000000;
    r[i] = -10000000000000;
    scanf("%d%d", &k, &b);
    e[i] = q[i] = (long long)k * x1 + b;
    w[i] = (long long)k * x2 + b;
  }
  sort(e + 1, e + n + 1);
  for (int i = 1; i <= n; i++) {
    aa = lower_bound(e + 1, e + n + 1, q[i]) - e - 1;
    as = ma1(aa);
    if (as > w[i]) {
      printf("YES\n");
      return 0;
    }
    ma0(++aa, w[i]);
    aa = n - aa;
    as = mi1(aa);
    if (as < w[i]) {
      printf("YES\n");
      return 0;
    }
    mi0(aa + 1, w[i]);
  }
  printf("NO\n");
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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, x1, x2, k, b, aa;
long long as, q[100009], w[100009], e[100009], r[100009], t[100009];
void ma0(int x, long long a) {
  for (; x <= n; x += x & -x) {
    if (a > r[x]) r[x] = a;
  }
}
long long ma1(int x) {
  long long a = -10000000000000;
  for (; x > 0; x -= x & -x)
    if (r[x] > a) a = r[x];
  return a;
}
void mi0(int x, long long a) {
  for (; x <= n; x += x & -x) {
    if (a < t[x]) t[x] = a;
  }
}
long long mi1(int x) {
  long long a = 10000000000000;
  for (; x > 0; x -= x & -x)
    if (t[x] < a) a = t[x];
  return a;
}
int main() {
  scanf("%d%d%d", &n, &x1, &x2);
  for (int i = 1; i <= n; i++) {
    t[i] = 10000000000000;
    r[i] = -10000000000000;
    scanf("%d%d", &k, &b);
    e[i] = q[i] = (long long)k * x1 + b;
    w[i] = (long long)k * x2 + b;
  }
  sort(e + 1, e + n + 1);
  for (int i = 1; i <= n; i++) {
    aa = lower_bound(e + 1, e + n + 1, q[i]) - e - 1;
    as = ma1(aa);
    if (as > w[i]) {
      printf("YES\n");
      return 0;
    }
    ma0(++aa, w[i]);
    aa = n - aa;
    as = mi1(aa);
    if (as < w[i]) {
      printf("YES\n");
      return 0;
    }
    mi0(aa + 1, w[i]);
  }
  printf("NO\n");
  return 0;
}

```