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

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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 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#include <bits/stdc++.h>
using namespace std;
FILE* in;
FILE* out;
const int MAX = 200004;
const int MM = 10002;
const int INF = 1000000001;
int n;
double a[MAX][2];
int idx[MM];
void update(double s1, double r1, double s2, double r2, double& lower,
            double& upper) {
  if (r1 == r2) {
    if (s1 < s2) upper = lower - 1;
    return;
  }
  if (r2 < r1) {
    lower = max(lower, (r1 * r2 * (s1 - s2)) / (s1 * s2 * (r2 - r1)));
  } else {
    upper = min(upper, (r1 * r2 * (s1 - s2)) / (s1 * s2 * (r2 - r1)));
  }
}
int main(void) {
  in = stdin;
  out = stdout;
  fscanf(in, "%d", &n);
  for (int i = 0; i < n; i++) fscanf(in, "%lf %lf", &a[i][0], &a[i][1]);
  memset(idx, -1, sizeof(idx));
  for (int i = 0; i < n; i++) {
    if (idx[(int)a[i][0]] == -1)
      idx[(int)a[i][0]] = i;
    else {
      if (a[idx[(int)a[i][0]]][1] < a[i][1]) idx[(int)a[i][0]] = i;
    }
  }
  vector<int> candid;
  for (int i = MM - 1; i >= 0; i--)
    if (idx[i] != -1) candid.push_back(idx[i]);
  vector<int> ans(1, candid[0]);
  long long lastS = a[candid[0]][0], lastR = a[candid[0]][1];
  for (int i = 1; i < (int)candid.size(); i++) {
    double s1 = a[candid[i]][0], r1 = a[candid[i]][1];
    double lower = 0.0, upper = INF;
    update(s1, r1, lastS, lastR, lower, upper);
    for (int c = i + 1; c < (int)candid.size(); c++) {
      update(s1, r1, a[candid[c]][0], a[candid[c]][1], lower, upper);
      if (upper < lower) break;
    }
    if (upper < lower) continue;
    ans.push_back(candid[i]);
    lastS = s1, lastR = r1;
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < (int)ans.size(); i++)
    fprintf(out, "%d%c", ans[i] + 1, i + 1 == (int)ans.size() ? '\n' : ' ');
  return 0;
}
