No matter what trouble you're in, don't be afraid, but face it with a smile.

I've made another billion dollars!

— Boboniu

Boboniu has issued his currencies, named Bobo Yuan. Bobo Yuan (BBY) is a series of currencies. Boboniu gives each of them a positive integer identifier, such as BBY-1, BBY-2, etc.

Boboniu has a BBY collection. His collection looks like a sequence. For example:

<image>

We can use sequence a=[1,2,3,3,2,1,4,4,1] of length n=9 to denote it.

Now Boboniu wants to fold his collection. You can imagine that Boboniu stick his collection to a long piece of paper and fold it between currencies:

<image>

Boboniu will only fold the same identifier of currencies together. In other words, if a_i is folded over a_j (1≤ i,j≤ n), then a_i=a_j must hold. Boboniu doesn't care if you follow this rule in the process of folding. But once it is finished, the rule should be obeyed.

A formal definition of fold is described in notes.

According to the picture above, you can fold a two times. In fact, you can fold a=[1,2,3,3,2,1,4,4,1] at most two times. So the maximum number of folds of it is 2.

As an international fan of Boboniu, you're asked to calculate the maximum number of folds. 

You're given a sequence a of length n, for each i (1≤ i≤ n), you need to calculate the maximum number of folds of [a_1,a_2,…,a_i].

Input

The first line contains an integer n (1≤ n≤ 10^5).

The second line contains n integers a_1,a_2,…,a_n (1≤ a_i≤ n).

Output

Print n integers. The i-th of them should be equal to the maximum number of folds of [a_1,a_2,…,a_i].

Examples

Input


9
1 2 3 3 2 1 4 4 1


Output


0 0 0 1 1 1 1 2 2


Input


9
1 2 2 2 2 1 1 2 2


Output


0 0 1 2 3 3 4 4 5


Input


15
1 2 3 4 5 5 4 3 2 2 3 4 4 3 6


Output


0 0 0 0 0 1 1 1 1 2 2 2 3 3 0


Input


50
1 2 4 6 6 4 2 1 3 5 5 3 1 2 4 4 2 1 3 3 1 2 2 1 1 1 2 4 6 6 4 2 1 3 5 5 3 1 2 4 4 2 1 3 3 1 2 2 1 1


Output


0 0 0 0 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 4 4 4 5 5 6 7 3 3 3 4 4 4 4 3 3 4 4 4 4 4 5 5 5 5 6 6 6 7 7 8

Note

Formally, for a sequence a of length n, let's define the folding sequence as a sequence b of length n such that:

  * b_i (1≤ i≤ n) is either 1 or -1. 
  * Let p(i)=[b_i=1]+∑_{j=1}^{i-1}b_j. For all 1≤ i<j≤ n, if p(i)=p(j), then a_i should be equal to a_j. 



([A] is the value of boolean expression A. i. e. [A]=1 if A is true, else [A]=0).

Now we define the number of folds of b as f(b)=∑_{i=1}^{n-1}[b_i≠ b_{i+1}].

The maximum number of folds of a is F(a)=max\{ f(b)∣ b  is a folding sequence of a \}.
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1, c = getchar();
  while (c < 48) c == '-' && (f = -1), c = getchar();
  while (c > 47) x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const int MAXN = 100005;
std::vector<int> adj[MAXN];
int s[MAXN], c[MAXN], q[MAXN];
int n, cnt;
inline bool check(int l, int r) {
  for (int x : adj[r])
    if (x == r - l + 1) return 1;
  return 0;
}
int main(int argc, char const *argv[]) {
  n = read();
  for (int i = q[0] = 1, x = 1; i <= n; ++x, ++i) {
    s[x] = read();
    adj[x].clear();
    if (x != 1) {
      if (s[x] == s[x - 1]) adj[x].push_back(2);
      for (int y : adj[x - 1])
        if (s[x - y - 1] == s[x]) adj[x].push_back(y + 2);
    }
    function<bool(void)> fes = [x]() {
      return check(x - adj[x][0] / 2 - adj[x][0] + 1, x - adj[x][0] / 2);
    };
    if (!adj[x].empty() && fes())
      x -= adj[x][0], cnt += 2;
    else {
      q[x] = q[x - 1];
      c[x] = c[x - 1];
      if (check(q[x], x)) q[x] += (x - q[x] + 1) / 2, c[x]++;
    }
    int ans = cnt + c[x], p = x;
    while (!adj[p].empty() && p - adj[p][0] / 2 >= q[x])
      p -= adj[p][0] / 2, ans++;
    printf("%d%c", ans, " \n"[i == n]);
  }
  return 0;
}
