#include <bits/stdc++.h>
using namespace std;
template <class T>
bool scan_d(T &ret) {
  char c;
  int sgn;
  T bit = 0.1;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && c != '.' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  if (c == ' ' || c == '\n') {
    ret *= sgn;
    return 1;
  }
  while (c = getchar(), c >= '0' && c <= '9') ret += (c - '0') * bit, bit /= 10;
  ret *= sgn;
  return 1;
}
inline void out(int x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
struct point {
  long long a;
  long long b;
};
point p[100010];
bool cmp(point a, point b) {
  if (a.a == b.a) return a.b < b.b;
  return a.a < b.a;
}
int main() {
  int n;
  scan_d(n);
  int i;
  long long x1, x2, k, b;
  scan_d(x1);
  scan_d(x2);
  for (i = 0; i < n; i++) {
    scan_d(k);
    scan_d(b);
    p[i].a = k * x1 + b;
    p[i].b = k * x2 + b;
  }
  sort(p, p + n, cmp);
  for (i = 1; i < n; i++) {
    if (p[i - 1].a != p[i].a && p[i - 1].b > p[i].b) {
      printf("yes");
      return 0;
    }
  }
  printf("no");
  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;
template <class T>
bool scan_d(T &ret) {
  char c;
  int sgn;
  T bit = 0.1;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && c != '.' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  if (c == ' ' || c == '\n') {
    ret *= sgn;
    return 1;
  }
  while (c = getchar(), c >= '0' && c <= '9') ret += (c - '0') * bit, bit /= 10;
  ret *= sgn;
  return 1;
}
inline void out(int x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
struct point {
  long long a;
  long long b;
};
point p[100010];
bool cmp(point a, point b) {
  if (a.a == b.a) return a.b < b.b;
  return a.a < b.a;
}
int main() {
  int n;
  scan_d(n);
  int i;
  long long x1, x2, k, b;
  scan_d(x1);
  scan_d(x2);
  for (i = 0; i < n; i++) {
    scan_d(k);
    scan_d(b);
    p[i].a = k * x1 + b;
    p[i].b = k * x2 + b;
  }
  sort(p, p + n, cmp);
  for (i = 1; i < n; i++) {
    if (p[i - 1].a != p[i].a && p[i - 1].b > p[i].b) {
      printf("yes");
      return 0;
    }
  }
  printf("no");
  return 0;
}

```