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: Accepted)

#define _USE_MATH_DEFINES
#include<iostream>
#include<vector>
#include<set>
#include<queue>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
#include<stack>
using namespace std;
#define MOD 1000000007
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define ll long long
#define ull unsigned long long
#define all(hoge) (hoge).begin(),(hoge).end()
typedef pair<ll, ll> P;
const long long INF = 1LL << 60;
typedef vector<ll> Array;
typedef vector<Array> Matrix;


//priority_queue<ll> max;//大きい順
//priority_queue<ll, Array, greater<ll>> min;//小さい順

/*firstについては昇順 secondについては降順
sort(all(wh), [&](P x, P y) {
	if (x.first == y.first)return x.second > y.second;
	return x.first < y.first;
});
*/

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}


//substr　文字列取り出し
//upper_bound ある値より大きい一番左のイテレータを返す、lowerは以上(setに対して使うとO(N)なので、setのメンバ関数を使う
//stoi


struct Edge {//グラフ
	ll to, cap, rev;
	Edge(ll _to, ll _cap, ll _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
void add_edge(Graph& G, ll from, ll to, ll cap,bool revFlag,ll revCap) {//最大フロー求める Ford-fulkerson
	G[from].push_back(Edge(to, cap, (ll)G[to].size()));
	if(revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));//最小カットの場合逆辺は0にする
}
ll max_flow_dfs(Graph & G, ll v, ll t, ll f, vector<bool> & used)
{
	if (v == t)
		return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); ++i) {
		Edge& e = G[v][i];
		if (!used[e.to] && e.cap > 0) {
			ll d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
			if (d > 0) {
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
//二分グラフの最大マッチングを求めたりも出来る　また二部グラフの最大独立集合は頂点数-最大マッチングのサイズ
ll max_flow(Graph & G, ll s, ll t)
{
	ll flow = 0;
	for (;;) {
		vector<bool> used(G.size());
		REP(i, used.size())used[i] = false;
		ll f = max_flow_dfs(G, s, t, INF, used);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
}
void BellmanFord(Graph& G, ll s, Array& d,Array &negative) {//O(|E||V|)
	d.resize(G.size());
	negative.resize(G.size());
	REP(i, d.size())d[i] = INF;
	REP(i, d.size())negative[i] = false;
	d[s] = 0;
	REP(k, G.size()-2) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
				}						   
			}
		}
	}
	REP(k, G.size() - 2) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
					negative[G[i][j].to] = true;
				}
				if(negative[i]==true)negative[G[i][j].to] = true;
			}
		}
	}
}
void Dijkstra(Graph& G, ll s, Array& d) {//O(|E|log|V|)
	d.resize(G.size());
	REP(i, d.size())d[i] = INF;
	d[s] = 0;
	priority_queue<P, vector<P>, greater<P>> q;
	q.push(make_pair(0, s));
	while (!q.empty()) {
		P a = q.top();
		q.pop();
		if (d[a.second] < a.first)continue;
		REP(i, G[a.second].size()) {
			Edge e = G[a.second][i];
			if (d[e.to] > d[a.second] + e.cap) {
				d[e.to] = d[a.second] + e.cap;
				q.push(make_pair(d[e.to], e.to));
			}
		}
	}
}
void WarshallFloyd(Graph& G, Matrix& d) {//O(V^3)
	d.resize(G.size());
	REP(i, d.size())d[i].resize(G.size());
	REP(i, d.size()) {
		REP(j, d[i].size()) {
			d[i][j] = INF;
		}
	}
	REP(i, G.size()) {
		REP(j, G[i].size()) {
			d[i][G[i][j].to] = G[i][j].cap;
		}
	}
	REP(i, G.size()) {
		REP(j, G.size()) {
			REP(k, G.size()) {
				chmin(d[j][k], d[j][i] + d[i][k]);
			}
		}
	}
}


class UnionFind {
	vector<int> data;
	ll num;
public:
	UnionFind(int size) : data(size, -1), num(size) { }
	bool unionSet(int x, int y) {//xとyの集合を統合する
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		num -= (x != y);
		return x != y;
	}
	bool findSet(int x, int y) {//xとyが同じ集合か返す
		return root(x) == root(y);
	}
	int root(int x) {//xのルートを返す
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {//xの集合のサイズを返す
		return -data[root(x)];
	}
	int numSet() {//集合の数を返す
		return num;
	}
};
class SumSegTree {
private:
	
	int _sum(int a, int b, int k, int l, int r) {
		if (r <= a || b <= l)return 0;    // 交差しない
		if (a <= l && r <= b)return dat[k];   // a,l,r,bの順で完全に含まれる
		else {
			int s1 = _sum(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
			int s2 = _sum(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
			return s1 + s2;
		}
	}
public:
	int n, height;
	vector<int> dat;

	// 初期化（_nは最大要素数）
	SumSegTree(int _n) {
		n = 1;
		height = 1;
		while (n < _n) {
			n *= 2;
			height++;
		}
		dat = vector<int>(2 * n - 1,0);
	}

	// 場所i(0-indexed)にxを足す
	void add(int i, int x) {
		i += n - 1; // i番目の葉ノードへ
		dat[i] += x;
		while (i > 0) { // 下から上がっていく
			i = (i - 1) / 2;
			dat[i] += x;
		}
	}

	// 区間[a,b)の総和。ノードk=[l,r)に着目している。
	int sum(int a, int b) {
		return _sum(a, b, 0, 0, n);
	}
};
class RmqTree {
private:

	ll _find(ll a, ll b, ll k, ll l, ll r) {
		if (r <= a || b <= l)return INF;    // 交差しない
		if (a <= l && r <= b)return dat[k];   // a,l,r,bの順で完全に含まれる
		else {
			ll s1 = _find(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
			ll s2 = _find(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
			return min(s1, s2);
		}
	}
public:
	ll n, height;
	vector<ll> dat;

	// 初期化（_nは最大要素数）
	RmqTree(ll _n) {
		n = 1;
		height = 1;
		while (n < _n) {
			n *= 2;
			height++;
		}
		dat = vector<ll>(2 * n - 1,INF);
	}

	// 場所i(0-indexed)をxにする
	void update(ll i, ll x) {
		i += n - 1; // i番目の葉ノードへ
		dat[i] = x;
		while (i > 0) { // 下から上がっていく
			i = (i - 1) / 2;
			dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
		}
	}

	// 区間[a,b)の最小値。ノードk=[l,r)に着目している。
	ll find(ll a, ll b) {
		return _find(a, b, 0, 0, n);
	}
};


//約数求める //約数
void divisor(ll n, vector<ll>& ret) {
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());

}

vector<ll> lis_fast(const vector<ll>& a) {//最長部分増加列
	const ll n = a.size();
	vector<ll> A(n, INT_MAX);
	vector<ll> id(n);
	for (int i = 0; i < n; ++i) {
		id[i] = distance(A.begin(), lower_bound(A.begin(), A.end(), a[i]));
		A[id[i]] = a[i];
	}
	ll m = *max_element(id.begin(), id.end());
	vector<ll> b(m + 1);
	for (int i = n - 1; i >= 0; --i)
		if (id[i] == m) b[m--] = a[i];
	return b;
}


ll ModPow(ll x, ll n) {
	ll res = 1LL;
	while (n > 0) {
		if (n & 1) res = res * x % MOD;
		x = x * x % MOD;
		n >>= 1;
	}
	return res;
}
//nCrとか
class Combination {
public:
	Array fact;
	Array inv;
	ll mod;
	ll mod_inv(ll x) {
		ll n = mod - 2LL;
		ll res = 1LL;
		while (n > 0) {
			if (n & 1) res = res * x % mod;
			x = x * x % mod;
			n >>= 1;
		}
		return res;
	}
	ll nCr(ll n, ll r) {
		return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;
	}
	ll nPr(ll n, ll r) {
		return (fact[n] * inv[n - r]) % mod;
	}
	ll nHr(ll n, ll r) {
		return nCr(r + n - 1, r);
	}
	Combination(ll n, ll _mod) {
		mod = _mod;
		fact.resize(n + 1);
		fact[0] = 1;
		REP(i, n) {
			fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
		}
		inv.resize(n + 1);
		REP(i, n + 1) {
			inv[i] = mod_inv(fact[i]);
		}
	}
};

bool compare_by_b(pair<ll, ll> a, pair<ll, ll> b) {//降順second
	if (a.second != b.second) {
		return a.second < b.second;
	}
	else {
		return a.first > b.first;
	}
}

bool compare_by_a(pair<ll, ll> a, pair<ll, ll> b) {//降順first
	if (a.first != b.first) {
		return a.first > b.first;
	}
	else {
		return a.second > b.second;
	}
}

ll gcd(ll m, ll n) {
	if (n == 0)return m;
	return gcd(n, m % n);
}//gcd

ll lcm(ll m, ll n) {
	return m / gcd(m, n) * n;
}


Matrix mIdentity(ll n) {
	Matrix A(n, Array(n));
	for (int i = 0; i < n; ++i) A[i][i] = 1;
	return A;
}

Matrix mMul(const Matrix& A, const Matrix& B) {
	Matrix C(A.size(), Array(B[0].size()));
	for (int i = 0; i < C.size(); ++i)
		for (int j = 0; j < C[i].size(); ++j)
			for (int k = 0; k < A[i].size(); ++k)
				(C[i][j] += (A[i][k]%MOD) * (B[k][j]%MOD))%=MOD;
	return C;
}
// O( n^3 log e )
Matrix mPow(const Matrix & A, ll e) {
	return e == 0 ? mIdentity(A.size()) : 
		e % 2 == 0 ? mPow(mMul(A, A), e / 2) : mMul(A, mPow(A, e - 1));
}




int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	ll n;
	cin >> n;
	vector<P> a(n);
	REP(i, n)cin >> a[i].first;
	REP(i, n)a[i].second = i;
	sort(all(a), [&](P x, P y) {
		if (x.first == y.first)return x.second < y.second;
		else return x.first > y.first;
	});
	Array ans(n);
	ll num = 1, total = a[0].first;
	ll now = a[0].second;

	for(int i=1;i<n;i++) {
		if (a[i].second > now) {
			total += a[i].first;
			num++;
		}
		else {
			ans[now] += total - a[i].first * num;
			num++;
			total = num * a[i].first;
			now = a[i].second;
		}
	}
	ans[now] += total;
	REP(i, n)cout << ans[i] << endl;
	

	return 0;
}