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)

#include <bits/stdc++.h>
#define exrep(i, a, b) for(long long i = a; i <= b; i++)
#define rep(i,n)for(long long i=0;i<(long long)(n);i++)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
 
const ll MOD=1e9+7;
const ll INF=1e18;
const int MAX=510000;
const double pi=acos(-1);
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
ll modpow(ll a, ll n){
  if(n==0)return 1;
  if(n%2==0){
    ll t=modpow(a,n/2);
    return t*t%MOD;
  }
  return a*modpow(a,n-1)%MOD;
}
ll modinv(ll a){
  return modpow(a,MOD-2);
}
long long fac[MAX], finv[MAX], inv[MAX];
 
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
struct UnionFind {
    vector<int> par; // 親ノード
    vector<int> rank; // ランク
 
    UnionFind(int n = 1) {
        init(n);
    }
 
    void init(int n = 1) {
        par.resize(n); rank.resize(n);
        for (int i = 0; i < n; ++i) par[i] = i, rank[i] = 0;
    }
 
    int root(int x) {
        if (par[x] == x) {
            return x;
        }
        else {
            int r = root(par[x]);
            return par[x] = r;
        }
    }
 
    bool issame(int x, int y) {
        return root(x) == root(y);
    }
 
    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        return true;
    }
};

    
  
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false); 
  COMinit();
  
  ll n;
  cin >> n;
  priority_queue<P>q;
  vector<ll>ans(n+1,0);
  rep(i,n){
    ll a;
    cin >> a;
    q.push(P(a,i+1));
  }
  ll idx=n;
  
  while(!q.empty()){
    ll cnt=0;
    auto p=q.top();
    P mp;
    while(!q.empty() && q.top().first==p.first){
      cnt++;
      mp=q.top();
      q.pop();
    }
    idx=min(idx,mp.second);
    
    ll diff;
    if(!q.empty())diff=mp.first-q.top().first;
    else diff=mp.first;
    ans[idx]+=diff*(n-q.size());
  }
  rep(i,n) cout << ans[i+1] << endl;
      
      
      
  
}