Tavas is a cheerleader in the new sports competition named "Pashmaks".

<image>

This competition consists of two part: swimming and then running. People will immediately start running R meters after they finished swimming exactly S meters. A winner is a such person that nobody else finishes running before him/her (there may be more than one winner).

Before the match starts, Tavas knows that there are n competitors registered for the match. Also, he knows that i-th person's swimming speed is si meters per second and his/her running speed is ri meters per second. Unfortunately, he doesn't know the values of R and S, but he knows that they are real numbers greater than 0.

As a cheerleader, Tavas wants to know who to cheer up. So, he wants to know all people that might win. We consider a competitor might win if and only if there are some values of R and S such that with these values, (s)he will be a winner.

Tavas isn't really familiar with programming, so he asked you to help him.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 2 × 105).

The next n lines contain the details of competitors. i-th line contains two integers si and ri (1 ≤ si, ri ≤ 104).

Output

In the first and the only line of output, print a sequence of numbers of possible winners in increasing order.

Examples

Input

3
1 3
2 2
3 1


Output

1 2 3 


Input

3
1 2
1 1
2 1


Output

1 3 
#include <bits/stdc++.h>
template <class T>
void Read(T &ret) {
  ret = 0;
  bool ok = 0, u = 0;
  for (;;) {
    int c = getchar();
    if (c >= '0' && c <= '9')
      ret = (ret << 3) + (ret << 1) + c - '0', ok = 1;
    else if (c == '-')
      u = 1;
    else if (ok) {
      if (u) ret *= -1;
      return;
    }
  }
}
long long pow_mod(long long p, long long n, long long mod) {
  long long ret = 1;
  for (; n; n >>= 1) {
    if (n & 1) ret = ret * p % mod;
    p = p * p % mod;
  }
  return ret;
}
template <class T>
bool chmin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool chmax(T &a, const T &b) {
  return b > a ? a = b, 1 : 0;
}
const int Max_N = 2e5 + 9;
const int Max_M = 1e4 + 9;
const double eps = 1e-24;
int sgn(double x) { return (x > eps) - (x < -eps); }
int _;
struct Point {
  int X, Y;
  double x, y;
  Point() {
    X = Y = 0;
    x = y = 0;
  }
  Point(double tx, double ty) {
    x = tx;
    y = ty;
  }
  Point(std::pair<int, int> tmp) {
    X = tmp.first;
    Y = tmp.second;
    x = 1.0 / X;
    y = 1.0 / Y;
  }
  std::pair<int, int> query() { return std::make_pair(X, Y); }
  bool operator<(const Point &t) const {
    if (sgn(x - t.x) == 0) {
      return sgn(y - t.y) < 0;
    }
    return sgn(x - t.x) < 0;
  }
} an[Max_N], bn[Max_N];
Point operator-(const Point &a, const Point &b) {
  return Point(a.x - b.x, a.y - b.y);
}
double operator^(const Point &a, const Point &b) {
  return a.x * b.y - a.y * b.x;
}
int ConvexHull(Point *p, int n, Point *ch) {
  std::sort(p, p + n);
  int r = 0;
  for (int i = 0; i < n; ++i) {
    while (r > 1 && sgn((ch[r - 1] - ch[r - 2]) ^ (p[i] - ch[r - 2])) <= 0) {
      --r;
    }
    ch[r++] = p[i];
  }
  int mid = r;
  for (int i = n - 2; i >= 0; --i) {
    while (r > mid && sgn((ch[r - 1] - ch[r - 2]) ^ (p[i] - ch[r - 2])) <= 0) {
      --r;
    }
    ch[r++] = p[i];
  }
  if (n > 1) --r;
  return r;
}
bool ans[Max_N];
std::vector<int> idx[Max_N];
std::map<std::pair<int, int>, int> map;
std::map<std::pair<int, int>, int>::iterator it;
int main() {
  int n, tot = 0;
  scanf("%d", &n);
  std::pair<int, int> tmp;
  for (int i = 0, x, y; i < n; ++i) {
    scanf("%d%d", &x, &y);
    tmp.first = x;
    tmp.second = y;
    it = map.find(tmp);
    if (it == map.end()) {
      map[tmp] = tot;
      idx[tot].push_back(i);
      ++tot;
    } else {
      idx[it->second].push_back(i);
    }
  }
  int m = 0;
  for (auto i : map) {
    an[m++] = Point(i.first);
  }
  int cnt = ConvexHull(an, m, bn);
  for (int i = 0; i < cnt; ++i) {
    if (i && sgn(bn[i].Y - bn[i - 1].Y) <= 0) break;
    for (auto j : idx[map[bn[i].query()]]) {
      ans[j] = true;
    }
  }
  for (int i = 0; i < n; ++i)
    if (ans[i]) {
      printf("%d ", i + 1);
    }
  return 0;
}
