#include <atcoder/lazysegtree.hpp>
#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
#define rep3(i, a, b, c) for (ll i = (a); i < (b); i += (c))
#define rep2(i, a, b) rep3(i, a, b, 1)
#define rep1(i, n) rep2(i, 0, n)
#define rep0(n) rep1(aaaaa, n)
#define ov4(a, b, c, d, name, ...) name
#define rep(...) ov4(__VA_ARGS__, rep3, rep2, rep1, rep0)(__VA_ARGS__)
#define per(i, a, b) for (ll i = (a) - 1; i >= (b); i--)
#define fore(e, v) for (auto &&e : v)
#define all(a) begin(a), end(a)
#define sz(a) (int)(size(a))
#define lb(v, x) (lower_bound(all(v), x) - begin(v))
#define eb emplace_back

template <typename T, typename S> bool chmin(T &a, const S &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T, typename S> bool chmax(T &a, const S &b) {
  return a < b ? a = b, 1 : 0;
}
const int INF = 1e9 + 100;
const ll INFL = 3e18 + 100;
#define i128 __int128_t
struct _ {
  _() { cin.tie(0)->sync_with_stdio(0), cout.tie(0); }
} __;

struct summax {
  ll s, m, l;
};

summax seg_op(summax a, summax b) {
  return summax{a.s + b.s, max(a.m, b.m), a.l + b.l};
}
summax seg_e() { return summax{0, -1, 0}; }
summax seg_map(ll v, summax a) {
  if (v < 0)
    return a;
  return summax{v * a.l, v, a.l};
}
ll seg_comp(ll f, ll g) {
  if (f < 0)
    return g;
  return f;
}
ll seg_id() { return -1; }

int main() {
  ll N;
  cin >> N;
  vi A(N);
  vector<vi> idx(N);
  rep(i, N) {
    cin >> A[i];
    if (A[i] < N)
      idx[A[i]].push_back(i);
  }
  vl cnt(N), ans(N + 1);
  atcoder::lazy_segtree<summax, seg_op, seg_e, ll, seg_map, seg_comp, seg_id>
      seg(N + 1);
  ll t = 0;
  rep(i, N) {
    t += i;
    seg.set(i, summax{i, i, 1});
  }
  rep(i, N) {
    int r1 = -1;
    idx[i].push_back(N);
    for (auto j : idx[i]) {
      auto f = [&](summax a) -> bool { return a.m < j; };
      auto r2 = seg.max_right(r1 + 1, f);
      seg.apply(r1 + 1, r2, j);
      r1 = r2;
    }
    ans[i] = seg.all_prod().s - t;
    // cerr<<"ans="<<ans[i]<<endl;
    // rep(i,N){
    //   cerr<<seg.get(i).m<<' ';
    // }
    // cerr<<endl;
  }
  ans[N] = N * (N + 1) / 2;
  per(i, N + 1, 1) { ans[i] -= ans[i - 1]; }
  rep(i, N + 1) cout << ans[i] << '\n';
}