/*

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;


#define num 200005
int input[num], input2[num];
int tmp[num], res[num];


void merge(int l,int mid,int r) {
    int i=l;
    int j=mid+1;
    int k=1;


    while (i <= mid && j <= r) {
        if (input[i] > input[j]) {
            
            //第i個比j大 由於i已經從小到大排過序了 那麼i+1到mid的也會比j大
            res[input[j]] += mid - i + 1;
            //cout<< input[i]<< " zz " << input[j] << " ";
            tmp[k++] = input[j++];
        } else {
            tmp[k++] = input[i++];
        }
    }
    while (i <= mid)
        tmp[k++] = input[i++];
    while (j <= r)
        tmp[k++] = input[j++];
    for (i = l, k = 1; i <= r; i++, k++) {
        input[i] = tmp[k];
    }
}


void merge_sort(int l,int r) {
    if(l < r) {
        int mid = (l + r) / 2;
        merge_sort(l, mid);
        //cout << l << " - " << mid << endl;
        merge_sort(mid + 1, r);
        //cout << mid + 1 << " -- " << r << endl;
        merge(l, mid, r);
    }
}


int main() {
    int n,i;
    scanf("%d", &n);
    for (i = 0; i < n; i++){
        scanf("%d", &input[i]);
        input2[i] = input[i];
    }


    merge_sort(0, n - 1);
    
    for (int i = 0; i < n - 1; ++i)
        printf("%d ", res[input2[i]]);
    printf("%d\n", res[input2[n - 1]]);
  
    return 0;

}
 

/*
[分析]線段樹解法，將問題轉化為查詢某個區間和問題。e.g.序列4 5 1 3 2

1.更新4時，更新4所在區間和+1，查詢[5,20w]的區間和即代表前序列中大於4的元素的個數。
2.同理
3.更新1時，即所有區間和都+1，然後再查詢[2, 20w]的區間和。
...
*/

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std;

#define N 200010

int sum[N*4];

void pushup(int rt) {
    sum[rt] = sum[rt*2] + sum[rt*2 + 1];
}

void build(int l,int r,int rt) {
    if (l==r) {
        sum[rt] = 0;
        return;
    }
    int m = (l + r) >> 1;
    build(l,m,rt*2);
    build(m + 1,r,rt * 2 + 1);
    pushup(rt);
}

void update(int p,int add,int l,int r,int rt) {
    if(l==r) {
        sum[rt] += add;
        return;
    }
    int m = (l + r) >> 1;
    if(p <= m) update(p,add,l,m,rt*2);
    else update(p,add,m + 1,r,rt * 2 + 1);

    pushup(rt);
}

int query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
        return sum[rt];
    }
    int m = (l + r) >> 1;
    int ans1 = 0, ans2 = 0;
    if (L <= m) ans1 += query(L,R,l,m,rt * 2);
    if (R > m) ans2 += query(L,R,m + 1,r,rt * 2 + 1);
    return ans1 + ans2;
}

int main() {
    //freopen("data/xianduanshu.txt","r",stdin);
    int n;
    scanf("%d",&n);
    build(1,200000,1);
    for(int i=1; i<=n; ++i) {
        int a;
        scanf("%d",&a);
        update(a,1,1,200000,1);
        //printf("rt = %d\n", sum[1]);
        printf("%d%c",query(a + 1,200000,1,200000,1), i==n?'\n':' ');//統計當前序列中大於a的元素的個數
    }
    return 0;
}