// Original C++ solution (slower):
/**       In the name of Allah, the Gracious, the Merciful.

*               Praise be to Allah, Lord of the Worlds.

*               **************************************

*

*    I have not failed, I have just found 1000 WAYS that won't work.

*                 __________________________________

*                  Author  : Abd El-Twab  M. Fakhry

*                 __________________________________

*/



#include <ext/rope>

#include <bits/stdc++.h>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/pb_ds/assoc_container.hpp>



#define  read(arr, l, r)     for(int i = (l); i < (r); cin >> arr[i++])

#define  print(arr, l, r)    for(int i = (l); i < (r); cout << arr[i++] << " \n"[i == r])

#define  rep(i, l, r)        for (int i = l; i < r; i++)

#define  repr(i, r, l)       for (int i = r; i > l; i--)

#define  trav(a, x)          for (auto & a : x)

#define  PQ(T)               priority_queue < T, vector <T> , greater <T> >

#define  Unique(n)           (n).erase(unique(all(n)), (n).end())

#define  endl                '\n'

#define  all(a)              a.begin() , a.end()

#define  NumberOfDigits(n)   ((int)log10(n) + 1)

#define  sz(a)               (int)a.size()

#define  rall(a)             a.rbegin(), a.rend()

#define  rsz(sz)             resize(sz)

#define  ins(x)              insert(x)

#define  debug(x)            cout << #x << " = " << (x) << '\n'

#define  LINE_ON             cout << "line Number = " << __LINE__ << endl

#define  RT(n)               return cout << (n) << endl, 0;

#define  clr(arr, x)         memset(arr, x, sizeof(arr))

#define  F                   first

#define  S                   second

#define  PB                  push_back

#define  EB                  emplace_back

#define  get(n)              find_by_order(n)

#define  indexOf(n)          order_of_key(n)

#define  isPowerOfTwo(x)     (x && !(x & (x - 1)))

#define  Swap(n, k)           n ^= k, k ^= n, n ^= k;



using namespace std;

using namespace __gnu_pbds;

using namespace __gnu_cxx;



template <class T>

using indexed_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update >;

typedef long long            ll;

typedef unsigned int         uint;

typedef unsigned long long   ull;



int Dx8[8] = { 1, -1, 0, 0, 1, 1, -1, -1 }, Dx4[4] = { 0, 0, 1,-1};

int Dy8[8] = { 0, 0, 1, -1, 1, -1, 1, -1 }, Dy4[4] = { 1,-1, 0, 0};



const double inf  = 1.0 / 0.0,      pi  = 2 * acos(0.0),   error = 1e-9;

const int    MxsZ = (int)2e6 + 6,   mod = (1 << 30) + 7,      oo = 0x3f3f3f3f;



ll Gcd(ll a, ll b)     { return b ? Gcd(b, a % b) : a; }

ll Lcm(ll a, ll b)     { return a * b / Gcd(a, b); }



void Fast()

{

    cin.sync_with_stdio(0);

    cin.tie(0);cout.tie(0);

    cin.exceptions(cin.failbit);

}



void File()

{

    freopen("input.in", "r", stdin);

    freopen("output.out", "w", stdout);

}



/**----------------------->>  Template  <<-----------------------**/



int n;

map <int, int> ma;



/**------------------>>  Initialization gap  <<------------------**/



int main()

{

    Fast();



    cin >> n; vector <int> v(n), temp; read(v, 0, n);

    rep(i, 0, n)

        ma[v[i]] = i + 1;

    for(auto it : ma) cout << it.S << " ";

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>



using namespace std;



struct Stu

{

    int num, rank;

    bool operator < (const Stu& x)const {

        return rank < x.rank;

    }

}stu[100010];

int n;



int main()

{

    scanf("%d", &n);

    for(int i = 0; i < n; i++)

    {

        scanf("%d", &stu[i].rank);

        stu[i].num = i + 1;

    }

    sort(stu, stu + n);

    printf("%d", stu[0].num);

    for(int i = 1; i < n; i++)printf( " %d", stu[i].num);

}

// Fastest known solution for this problem: