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 
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 

/*
ID: espr1t
TASK: 
KEYWORDS: 
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>

using namespace std;
FILE *in; FILE *out;

const int MAX = 200004;
const int MM = 10002;
const double EPS = 0.000000001;
const double INF = 1000000001.0;

int n;
double a[MAX][2];
int idx[MM];

void update(double s1, double r1, double s2, double r2, double& lower, double& upper) {
    if (fabs(r1 - r2) < EPS) {
        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;
//	in = fopen("C.in", "rt"); out = fopen("C.out", "wt");
	
	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;
	    }
	}
	
	// At the beginning S -> INF, R -> ZERO
	// At the end S -> ZERO, R -> INF
	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]);
	double 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 = EPS, upper = INF * 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;
}
