#include<bits/stdc++.h>

#define ri register signed

#define p(i) ++i

using namespace std;

namespace IO{

    char buf[1<<21],*p1=buf,*p2=buf;

    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++

    template<typename T>inline void read(T &x) {

        ri f=1;x=0;register char ch=gc();

        while(ch<'0'||ch>'9') {if (ch=='-') f=0;ch=gc();}

        while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}

        x=f?x:-x;

    }

}

using IO::read;

namespace nanfeng{

    #define FI FILE *IN

    #define FO FILE *OUT

    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}

    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}

    typedef double db;

    static const int N=3e5+7;

    int st[N],tp,pre[N],vis[N],n,mxa,mxb;

    db tk[N];

    struct node{int a,b,id;}mis[N];

    inline int operator<(const node &n1,const node &n2) {return n1.a==n2.a?n1.b>n2.b:n1.a>n2.a;}//每个相同的 a 只有最大的 b 有用

    inline db slope(node n1,node n2) {return ((db)n1.a*n2.a*(n2.b-n1.b))/((db)n1.b*n2.b*(n2.a-n1.a));}

    inline int main() {

        // FI=freopen("nanfeng.in","r",stdin);

        // FO=freopen("nanfeng.out","w",stdout);

        read(n);

        for (ri i(1);i<=n;p(i)) read(mis[i].a),read(mis[i].b),mis[i].id=i;

        sort(mis+1,mis+n+1);

        for (ri i(1);i<=n;p(i)) if(mxb<mis[i].b) mxb=mis[i].b,mxa=mis[i].a;

        st[p(tp)]=1;

        for (ri i(2);mxa<=mis[i].a&&i<=n;p(i)) {//mxa 防止右凸包也算进去

            if (mis[i].a==mis[st[tp]].a) {

                if (mis[i].b==mis[st[tp]].b)

                    pre[mis[i].id]=pre[mis[st[tp]].id],pre[mis[st[tp]].id]=mis[i].id;

                continue;

            }

            while(tp>1&&slope(mis[st[tp]],mis[i])<tk[st[tp]]) --tp;

            tk[i]=slope(mis[st[tp]],mis[i]),st[p(tp)]=i;

        }

        while(tp)

            for (ri i(mis[st[tp--]].id);i;i=pre[i]) vis[i]=1;

        for (ri i(1);i<=n;p(i)) if (vis[i]) printf("%d ",i);

        return 0;

    }

}

int main() {return nanfeng::main();}