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: Time Limit Exceeded)

//header
#define set_header 1
#ifdef set_header
    #include <bits/stdc++.h>
    #ifdef LOCAL
        #include "cxx-prettyprint-master/prettyprint.hpp"
        #define print(x) cout << x << endl
    #else
        #define print(...) 42
    #endif
    using namespace std;
    using ll = long long;
    #define INF 1'010'000'000'000'000'017LL
    #define mod 998244353LL
    #define eps 0.0001
    #define all(x) (x).begin(), (x).end()
    #define reverse(x) (x).rbegin(), (x).rend()
    #define FOR(i,a,b) for(ll i=(a);i<(b);++i)
    #define rep(i,n)  FOR(i,0,n)
    #define SZ(x) ((ll)(x).size())
    #define sum(x) accumulate(ALL(x), 0LL)//?
    #define pb(x) push_back(x)
    typedef pair < ll , ll >P;
    ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
    template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
    template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
    template <std::uint_fast64_t Modulus> class modint {
        using u64 = std::uint_fast64_t;

        public:
        u64 a;

        constexpr modint(const u64 x = 0) noexcept : a(x % Modulus) {}
        constexpr u64 &value() noexcept { return a; }
        constexpr const u64 &value() const noexcept { return a; }
        constexpr modint operator+(const modint rhs) const noexcept {
            return modint(*this) += rhs;
        }
        constexpr modint operator-(const modint rhs) const noexcept {
            return modint(*this) -= rhs;
        }
        constexpr modint operator*(const modint rhs) const noexcept {
            return modint(*this) *= rhs;
        }
        constexpr modint operator/(const modint rhs) const noexcept {
            return modint(*this) /= rhs;
        }
        constexpr modint &operator+=(const modint rhs) noexcept {
            a += rhs.a;
            if (a >= Modulus) {
            a -= Modulus;
            }
            return *this;
        }
        constexpr modint &operator-=(const modint rhs) noexcept {
            if (a < rhs.a) {
            a += Modulus;
            }
            a -= rhs.a;
            return *this;
        }
        constexpr modint &operator*=(const modint rhs) noexcept {
            a = a * rhs.a % Modulus;
            return *this;
        }
        constexpr modint &operator/=(modint rhs) noexcept {
            u64 exp = Modulus - 2;
            while (exp) {
            if (exp % 2) {
                *this *= rhs;
            }
            rhs *= rhs;
            exp /= 2;
            }
            return *this;
        }
        };
    template< typename T >
    T mod_pow(T x, T n, const T &p) {
        T ret = 1;
        while(n > 0) {
            if(n & 1) (ret *= x) %= p;
            (x *= x) %= p;
            n >>= 1;
        }
        return ret;
    }
    class range {private: struct I{int x;int operator*(){return x;}bool operator!=(I& lhs){return x<lhs.x;}void operator++(){++x;}};I i,n;
    public:range(int n):i({0}),n({n}){}range(int i,int n):i({i}),n({n}){}I& begin(){return i;}I& end(){return n;}};
    uint64_t my_rand(void) {
	static uint64_t x = 88172645463325252ULL;
	x = x ^ (x << 13); x = x ^ (x >> 7);
	return x = x ^ (x << 17);
}
#endif
//library
//main
int main() {
    ll N;
    cin >> N;
    vector<ll> a(N+1), ans(N);
    rep(i, N)cin >> a[i+1];
    auto iter = max_element(all(a));
    priority_queue<ll> que;
    for(auto i = iter; i<end(a); i++){
        que.push(*i);
    }
    ll cnt = 0;
    int index = distance(begin(a), iter);
    while(index>=1){
        auto next_iter = max_element(begin(a), iter);
        ll res = cnt*(*iter-*next_iter);
        for(auto i = next_iter; i<iter;i++){
            que.push(*i);
        }
        while(SZ(que) && que.top()>*next_iter){
            res+=que.top()-*next_iter;
            cnt+=1;
            que.pop();
        }
        ans[index-1] = res;
        iter = next_iter;
        index = distance(begin(a), iter);
    }
    rep(i, N)cout << ans[i] << endl;
    return 0;
}