#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chmax(T& a, const T& b) {
  return a < b ? (a = b, true) : false;
}
template <typename T>
bool chmin(T& a, const T& b) {
  return b < a ? (a = b, true) : false;
}
long long N, X1, X2;
const long long INF = 1e17;
int main() {
  scanf("%lld%lld%lld", &N, &X1, &X2);
  vector<long long> K(N), B(N);
  vector<pair<long long, long long> > Y(N);
  for (int i = (int)(0); i < (int)(N); i++) {
    scanf("%lld%lld", &K[i], &B[i]);
    Y[i] = make_pair(K[i] * X1 + B[i], i);
  }
  sort(Y.begin(), Y.end());
  long long mx = -INF;
  for (int i = 0; i < N;) {
    int j = i;
    while (j < N && (Y[j].first == Y[i].first)) {
      long long idx = Y[j].second, t = K[idx] * X2 + B[idx];
      if (t < mx) {
        printf("YES\n");
        return 0;
      }
      j++;
    }
    for (int k = (int)(i); k < (int)(j); k++) {
      long long idx = Y[k].second, t = K[idx] * X2 + B[idx];
      chmax(mx, t);
    }
    i = j;
  }
  printf("NO\n");
  return 0;
}

### Prompt

Create a solution in Cpp for 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 <typename T>
bool chmax(T& a, const T& b) {
  return a < b ? (a = b, true) : false;
}
template <typename T>
bool chmin(T& a, const T& b) {
  return b < a ? (a = b, true) : false;
}
long long N, X1, X2;
const long long INF = 1e17;
int main() {
  scanf("%lld%lld%lld", &N, &X1, &X2);
  vector<long long> K(N), B(N);
  vector<pair<long long, long long> > Y(N);
  for (int i = (int)(0); i < (int)(N); i++) {
    scanf("%lld%lld", &K[i], &B[i]);
    Y[i] = make_pair(K[i] * X1 + B[i], i);
  }
  sort(Y.begin(), Y.end());
  long long mx = -INF;
  for (int i = 0; i < N;) {
    int j = i;
    while (j < N && (Y[j].first == Y[i].first)) {
      long long idx = Y[j].second, t = K[idx] * X2 + B[idx];
      if (t < mx) {
        printf("YES\n");
        return 0;
      }
      j++;
    }
    for (int k = (int)(i); k < (int)(j); k++) {
      long long idx = Y[k].second, t = K[idx] * X2 + B[idx];
      chmax(mx, t);
    }
    i = j;
  }
  printf("NO\n");
  return 0;
}

```