Problem:
Score : 700 points

Problem Statement

Snuke loves constructing integer sequences.

There are N piles of stones, numbered 1 through N.
The pile numbered i consists of a_i stones.

Snuke will construct an integer sequence s of length Σa_i, as follows:

Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.

Select a pile with one or more stones remaining, and remove a stone from that pile.

If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.

We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?

Constraints

1 ≤ N ≤ 10^{5}

1 ≤ a_i ≤ 10^{9}

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.

Sample Input 1

2
1 2

Sample Output 1

2
1

The lexicographically smallest sequence is constructed as follows:

Since the pile with the largest number of stones remaining is pile 2, append 2 to the end of s. Then, remove a stone from pile 2.

Since the piles with the largest number of stones remaining are pile 1 and 2, append 1 to the end of s (we take the smallest index). Then, remove a stone from pile 2.

Since the pile with the largest number of stones remaining is pile 1, append 1 to the end of s. Then, remove a stone from pile 1.

The resulting sequence is (2,1,1). In this sequence, 1 occurs twice, and 2 occurs once.

Sample Input 2

10
1 2 1 3 2 4 2 5 8 1

Sample Output 2

10
7
0
4
0
3
0
2
3
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <iomanip>
#include <math.h>
#include <string.h>
#include <cstdio>
#include <tuple>
using namespace std; using ll = long long; using ld = long double;  using pll = pair<ll, ll>;
using vl = vector<ll>; using vll = vector<vl>; using vpll = vector<pll>;
using vs = vector<string>; using tll = tuple<ll, ll, ll>; using vtll = vector<tll>;
const ld PI = 3.1415926535897932;
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define rep1(n) for(ll i=0;i<n;++i)
#define rep2(i,n) for(ll i=0;i<n;++i)
#define rep3(i,a,b) for(ll i=a;i<b;++i)
#define rep4(i,a,b,c) for(ll i=a;i<b;i+=c)
#define rep(...) overload4(__VA_ARGS__,rep4,rep3,rep2,rep1)(__VA_ARGS__)
#define mp make_pair
#define mt make_tuple
void in() {}
void debug_out() { cerr << endl; }
template<typename Head, typename... Tail>
void debug_out(Head h, Tail... t) {
	cerr << " " << h;
	if (sizeof...(t) > 0) cerr << " :";
	debug_out(t...);
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> vec) {
	for (size_t i = 0; i < vec.size(); i++)os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
ll ctoi(char c) {
	switch (c) {
	case '0': return 0; case '1': return 1; case '2': return 2;
	case '3': return 3; case '4': return 4; case '5': return 5;
	case '6': return 6; case '7': return 7; case '8': return 8;
	case '9': return 9;  default: return 0;
	}
}
bool pairCompare(const pll firstElof, pll secondElof)
{
	return firstElof.first > secondElof.first;
}
//**Snippetリスト**//
//rep, vin, all, iteLoop, bitSearch, bitList, nod//
//gcdlcm, isPrime, eratos, primeFactorize, Npow, combination, divisor, modinv,  digitDP//
//dfs, bfs, dijkstra, WarshallFloyd, BellmanFord, UnionFind, Kruskal ,RMQ, LCA//
ll i, j, k, l; ll N, M, K, H, W, L, X, Y, Z, R, Q;
ll MOD = 1000000007; ll INF = 1LL << 60; ll ans = 0; ll z = 0, o = 1;
vl flag; vector<vpll> path, path2; vl D1, D2;
//***********//
int main() {
	cin >> N;
	vpll A(N);
	for (i = 0; i < N; i++) {
		cin >> A[i].first;
		A[i].second = i;
	}
	vl Ans(N);
	sort(A.begin(), A.end());
	reverse(A.begin(), A.end());
	ll count = 1;
	ll pos = A[0].second;
	ll prev = A[0].first;
	ll now = 0;
	for (i = 0; i < N-1; i++) {
		if (A[i].first == A[i+1].first) {
			pos = A[i+1].second;
			count++;
		}
		else {
			now = A[i + 1].first;
			Ans[pos] += count * (prev-now);
			prev = now;
			pos = A[i+1].second;
			count++;
		}
	}
	Ans[pos] += count * prev;
	for (i = 0; i < N; i++) {
		cout << Ans[i] << endl;
	}
}