#include <bits/stdc++.h>
using namespace std;
unordered_map<int, int> ser[100010];
vector<int> LP[100010], RP[100010];
int dfs(int L, int R) {
  if (ser[R].count(L)) {
    return ser[R][L];
  }
  int ans = 0;
  for (int i = 0; i < LP[L].size() && L + 2 * LP[L][i] - 1 <= R; i++)
    ans = max(ans, dfs(L + LP[L][i], R) + 1);
  for (int i = 0; i < RP[R].size() && R - 2 * RP[R][i] + 1 >= L; i++)
    ans = max(ans, dfs(L, R - RP[R][i]) + 1);
  return ser[R][L] = ans;
}
int n, m, S[100010];
int main() {
  int ch, sumz = 0;
  scanf("%d", &n);
  while (n--) {
    scanf("%d", &ch);
    if (ch == S[m]) {
      RP[m + 1].push_back(1);
    }
    for (int i = 0; i < RP[m].size(); i++)
      if (S[m - RP[m][i] * 2] == ch) {
        RP[m + 1].push_back(RP[m][i] + 1);
      }
    int zlen = 0;
    for (int i = RP[m + 1].size() - 1; i >= 0; i--) {
      int pos = m + 1 - RP[m + 1][i];
      int x = lower_bound(RP[pos].begin(), RP[pos].end(), RP[m + 1][i]) -
              RP[pos].begin();
      if (x != RP[pos].size() && RP[pos][x] == RP[m + 1][i]) {
        zlen = RP[m + 1][i];
        break;
      }
    }
    S[++m] = ch;
    for (int i = 0; i < RP[m].size(); i++) {
      LP[m - 2 * RP[m][i] + 1].push_back(RP[m][i]);
    }
    if (zlen) {
      for (int ti = 0; ti < 2 * zlen; ti++) {
        for (int i = 0; i < RP[m].size(); i++)
          LP[m - 2 * RP[m][i] + 1].pop_back();
        RP[m].clear();
        ser[m].clear();
        m--;
      }
      ++sumz;
    }
    printf("%d ", sumz * 2 + dfs(1, m));
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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 \}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
unordered_map<int, int> ser[100010];
vector<int> LP[100010], RP[100010];
int dfs(int L, int R) {
  if (ser[R].count(L)) {
    return ser[R][L];
  }
  int ans = 0;
  for (int i = 0; i < LP[L].size() && L + 2 * LP[L][i] - 1 <= R; i++)
    ans = max(ans, dfs(L + LP[L][i], R) + 1);
  for (int i = 0; i < RP[R].size() && R - 2 * RP[R][i] + 1 >= L; i++)
    ans = max(ans, dfs(L, R - RP[R][i]) + 1);
  return ser[R][L] = ans;
}
int n, m, S[100010];
int main() {
  int ch, sumz = 0;
  scanf("%d", &n);
  while (n--) {
    scanf("%d", &ch);
    if (ch == S[m]) {
      RP[m + 1].push_back(1);
    }
    for (int i = 0; i < RP[m].size(); i++)
      if (S[m - RP[m][i] * 2] == ch) {
        RP[m + 1].push_back(RP[m][i] + 1);
      }
    int zlen = 0;
    for (int i = RP[m + 1].size() - 1; i >= 0; i--) {
      int pos = m + 1 - RP[m + 1][i];
      int x = lower_bound(RP[pos].begin(), RP[pos].end(), RP[m + 1][i]) -
              RP[pos].begin();
      if (x != RP[pos].size() && RP[pos][x] == RP[m + 1][i]) {
        zlen = RP[m + 1][i];
        break;
      }
    }
    S[++m] = ch;
    for (int i = 0; i < RP[m].size(); i++) {
      LP[m - 2 * RP[m][i] + 1].push_back(RP[m][i]);
    }
    if (zlen) {
      for (int ti = 0; ti < 2 * zlen; ti++) {
        for (int i = 0; i < RP[m].size(); i++)
          LP[m - 2 * RP[m][i] + 1].pop_back();
        RP[m].clear();
        ser[m].clear();
        m--;
      }
      ++sumz;
    }
    printf("%d ", sumz * 2 + dfs(1, m));
  }
  return 0;
}

```