Algorithm


B. Queue
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

There are n walruses standing in a queue in an airport. They are numbered starting from the queue's tail: the 1-st walrus stands at the end of the queue and the n-th walrus stands at the beginning of the queue. The i-th walrus has the age equal to ai.

The i-th walrus becomes displeased if there's a younger walrus standing in front of him, that is, if exists such j (i < j), that ai > aj. The displeasure of the i-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the i-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.

The airport manager asked you to count for each of n walruses in the queue his displeasure.

Input

The first line contains an integer n (2 ≤ n ≤ 105) — the number of walruses in the queue. The second line contains integers ai (1 ≤ ai ≤ 109).

Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be strictly younger than the other one.

Output

Print n numbers: if the i-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the i-th walrus's displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.

Examples
input
Copy
6
10 8 5 3 50 45
output
Copy
2 1 0 -1 0 -1 
input
Copy
7
10 4 6 3 2 8 15
output
Copy
4 2 1 0 -1 -1 -1 
input
Copy
5
10 3 1 10 11
output
Copy
1 0 -1 -1 -1 

 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>

int main(){

    long n; scanf("%ld", &n);
    std::vector<std::pair<long, long> > a(n); 
    for(long p = 0; p < n; p++){scanf("%ld", &a[p].first); a[p].second = p;}
    sort(a.begin(), a.end());
    long mx(-1);
    std::vector<long> f(n); 
    for(long p = 0; p < n; p++){
        f[a[p].second] = (a[p].second > mx) ? (-1) : (mx - a[p].second - 1);
        mx = (mx > a[p].second) ? mx : a[p].second;
    }

    for(long p = 0; p < n; p++){printf("%ld ", f[p]);}
    puts("");

    return 0;
}
Copy The Code & Try With Live Editor

Input

x
+
cmd
6
10 8 5 3 50 45

Output

x
+
cmd
2 1 0 -1 0 -1
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Demonstration


Codeforces Solution-B. Queue-Solution in C, C++, Java, Python

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