Ildar is the algorithm teacher of William and Harris. Today, Ildar is teaching Cartesian Tree. However, Harris is sick, so Ildar is only teaching William.

A cartesian tree is a rooted tree, that can be constructed from a sequence of distinct integers. We build the cartesian tree as follows:

  1. If the sequence is empty, return an empty tree;
  2. Let the position of the maximum element be x;
  3. Remove element on the position x from the sequence and break it into the left part and the right part (which might be empty) (not actually removing it, just taking it away temporarily);
  4. Build cartesian tree for each part;
  5. Create a new vertex for the element, that was on the position x which will serve as the root of the new tree. Then, for the root of the left part and right part, if exists, will become the children for this vertex;
  6. Return the tree we have gotten.



For example, this is the cartesian tree for the sequence 4, 2, 7, 3, 5, 6, 1:

<image>

After teaching what the cartesian tree is, Ildar has assigned homework. He starts with an empty sequence a.

In the i-th round, he inserts an element with value i somewhere in a. Then, he asks a question: what is the sum of the sizes of the subtrees for every node in the cartesian tree for the current sequence a?

Node v is in the node u subtree if and only if v = u or v is in the subtree of one of the vertex u children. The size of the subtree of node u is the number of nodes v such that v is in the subtree of u.

Ildar will do n rounds in total. The homework is the sequence of answers to the n questions.

The next day, Ildar told Harris that he has to complete the homework as well. Harris obtained the final state of the sequence a from William. However, he has no idea how to find the answers to the n questions. Help Harris!

Input

The first line contains a single integer n (1 ≤ n ≤ 150000).

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guarenteed that each integer from 1 to n appears in the sequence exactly once.

Output

Print n lines, i-th line should contain a single integer — the answer to the i-th question.

Examples

Input


5
2 4 1 5 3


Output


1
3
6
8
11


Input


6
1 2 4 5 6 3


Output


1
3
6
8
12
17

Note

After the first round, the sequence is 1. The tree is

<image>

The answer is 1.

After the second round, the sequence is 2, 1. The tree is

<image>

The answer is 2+1=3.

After the third round, the sequence is 2, 1, 3. The tree is

<image>

The answer is 2+1+3=6.

After the fourth round, the sequence is 2, 4, 1, 3. The tree is

<image>

The answer is 1+4+1+2=8.

After the fifth round, the sequence is 2, 4, 1, 5, 3. The tree is

<image>

The answer is 1+3+1+5+1=11.
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 150000;
struct SegtreeBeatsMin {
  int n, lim;
  vector<int> maxval, maxval2;
  vector<int> maxcnt;
  vector<bool> ischanged;
  vector<int> changes;
  vector<int> changedby;
  void logchange(int val, int by) {
    if (val == -1) return;
    if (!ischanged[val]) ischanged[val] = true, changes.push_back(val);
    changedby[val] += by;
  }
  void sapplymin(int x, int val, bool log) {
    if (val >= maxval[x]) return;
    assert(val > maxval2[x]);
    if (log) logchange(maxval[x], -maxcnt[x]);
    maxval[x] = val;
    if (log) logchange(maxval[x], +maxcnt[x]);
  }
  void spush(int x, int l, int m, int r) {
    sapplymin(x + 1, maxval[x], false);
    sapplymin(x + 2 * (m - l + 1), maxval[x], false);
  }
  void spull(int x, int l, int m, int r) {
    maxval[x] = max(maxval[x + 1], maxval[x + 2 * (m - l + 1)]);
    maxval2[x] =
        max(maxval[x + 1] != maxval[x] ? maxval[x + 1] : maxval2[x + 1],
            maxval[x + 2 * (m - l + 1)] != maxval[x]
                ? maxval[x + 2 * (m - l + 1)]
                : maxval2[x + 2 * (m - l + 1)]);
    maxcnt[x] =
        (maxval[x + 1] == maxval[x] ? maxcnt[x + 1] : 0) +
        (maxval[x + 2 * (m - l + 1)] == maxval[x] ? maxcnt[x + 2 * (m - l + 1)]
                                                  : 0);
  }
  void sinit(int x, int l, int r) {
    if (l == r) {
      maxval[x] = -1, maxval2[x] = INT_MIN, maxcnt[x] = 1;
    } else {
      int m = l + (r - l) / 2;
      sinit(x + 1, l, m);
      sinit(x + 2 * (m - l + 1), m + 1, r);
      spull(x, l, m, r);
    }
  }
  void sset(int x, int l, int r, int IDX, int VAL) {
    if (l == r) {
      logchange(maxval[x], -maxcnt[x]);
      maxval[x] = VAL, maxval2[x] = INT_MIN, maxcnt[x] = 1;
      logchange(maxval[x], +maxcnt[x]);
    } else {
      int m = l + (r - l) / 2;
      spush(x, l, m, r);
      if (IDX <= m)
        sset(x + 1, l, m, IDX, VAL);
      else
        sset(x + 2 * (m - l + 1), m + 1, r, IDX, VAL);
      spull(x, l, m, r);
    }
  }
  void smin(int x, int l, int r, int L, int R, int VAL) {
    if (VAL >= maxval[x]) return;
    if (L <= l && r <= R && VAL > maxval2[x]) {
      sapplymin(x, VAL, true);
    } else {
      int m = l + (r - l) / 2;
      spush(x, l, m, r);
      if (L <= m) smin(x + 1, l, m, L, R, VAL);
      if (m + 1 <= R) smin(x + 2 * (m - l + 1), m + 1, r, L, R, VAL);
      spull(x, l, m, r);
    }
  }
  void init(int _n, int _lim) {
    n = _n, lim = _lim;
    maxval = maxval2 = vector<int>(2 * n - 1);
    maxcnt = vector<int>(2 * n - 1);
    sinit(0, 0, n - 1);
    ischanged = vector<bool>(lim, false);
    changes = vector<int>();
    changedby = vector<int>(lim, 0);
  }
  void set(int idx, int val) { sset(0, 0, n - 1, idx, val); }
  void minrange(int lidx, int ridx, int val) {
    smin(0, 0, n - 1, lidx, ridx, val);
  }
  void reset() {
    for (int i = (0); i < (((int)(changes).size())); ++i) {
      int val = changes[i];
      ischanged[val] = false;
      changedby[val] = 0;
    }
    changes.clear();
  }
};
struct BIT {
  int n;
  vector<long long> bit;
  long long sum;
  void init(int _n) {
    n = _n;
    bit = vector<long long>(n, 0);
    sum = 0;
  }
  void add(int idx, long long val) {
    sum += val;
    while (idx < n) {
      bit[idx] += val;
      idx |= idx + 1;
    }
  }
  long long prefixsum(int idx) {
    long long ret = 0;
    while (idx >= 0) {
      ret += bit[idx];
      idx = (idx & (idx + 1)) - 1;
    }
    return ret;
  }
  long long suffixsum(int idx) { return sum - prefixsum(idx - 1); }
};
int n;
int val[MAXN];
long long ans[MAXN];
int pos[MAXN];
SegtreeBeatsMin leftseg;
set<int> leftset;
BIT alivebit;
set<int> aliveset;
BIT freq;
void solvermax() {
  leftseg.init(n, n);
  leftset.clear();
  alivebit.init(n);
  aliveset.clear();
  freq.init(n);
  long long sum = 0;
  int fst = -1, lst = -1;
  for (int v = (0); v < (n); ++v) {
    int idx = pos[v];
    alivebit.add(idx, 1);
    aliveset.insert(idx);
    sum += freq.suffixsum(idx);
    if (fst == -1 || idx < fst) {
      fst = idx;
      if (fst != 0)
        freq.add(fst - 1, -1), sum -= alivebit.prefixsum(fst - 1),
            leftset.insert(fst);
      if (lst == -1) lst = idx;
    } else {
      auto it = aliveset.find(idx);
      int prv = *prev(it);
      leftseg.minrange(0, prv, prv);
      if (idx > lst) {
        lst = idx;
      } else {
        int nxt = *next(it);
        leftseg.set(nxt, lst);
        if (leftset.find(nxt) == leftset.end())
          freq.add(nxt - 1, -1), sum -= alivebit.prefixsum(nxt - 1),
              leftset.insert(nxt);
      }
    }
    leftseg.set(fst, lst);
    for (int i = (0); i < (((int)(leftseg.changes).size())); ++i) {
      int val = leftseg.changes[i], by = leftseg.changedby[val];
      freq.add(val, by), sum += (long long)by * alivebit.prefixsum(val);
    }
    leftseg.reset();
    ans[v] += sum;
  }
}
void solve() {
  for (int i = (0); i < (n); ++i) pos[val[i]] = i;
  for (int i = (0); i < (n); ++i) ans[i] = 0;
  for (int kind = (0); kind < (2); ++kind) {
    solvermax();
    for (int i = (0); i < (n); ++i) pos[i] = n - pos[i] - 1;
  }
  for (int i = (0); i < (n); ++i) ans[i] -= i + 1;
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &val[i]), --val[i];
  solve();
  for (int i = (0); i < (n); ++i) printf("%lld\n", ans[i]);
}
int main() {
  run();
  return 0;
}
