#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
int n, k, w[N];
vector<int> vl[N], vr[N];
unordered_map<int, int> F[N];
inline int dfs(int l, int r) {
  if (F[r].count(l)) return F[r][l];
  int ans = 0;
  for (int i = 0; i < vl[l].size(); i++)
    if (l + vl[l][i] <= r) ans = max(ans, dfs(l + (vl[l][i] + 1) / 2, r) + 1);
  for (int i = 0; i < vr[r].size(); i++)
    if (r - vr[r][i] >= l) ans = max(ans, dfs(l, r - (vr[r][i] + 1) / 2) + 1);
  F[r][l] = ans;
  return ans;
}
int main() {
  cin >> n;
  int x, ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x), w[++k] = x;
    if (w[k] == w[k - 1]) vr[k].push_back(1), vl[k - 1].push_back(1);
    for (int j = 0; j < vr[k - 1].size(); j++) {
      int x = k - 2 - vr[k - 1][j];
      if (w[x] == w[k])
        vl[x].push_back(vr[k - 1][j] + 2), vr[k].push_back(vr[k - 1][j] + 2);
    }
    if (vr[k].size() == 1) {
      int p = k - (vr[k][0] + 1) / 2;
      bool fl = 0;
      for (int j = 0; j < vr[p].size(); j++)
        if (vr[p][j] == vr[k][0]) fl = 1;
      if (fl) {
        p = k - vr[k][0];
        while (k >= p) {
          for (int j = 0; j < vr[k].size(); j++)
            vl[k - vr[k][j]].resize(vl[k - vr[k][j]].size() - 1);
          vr[k].resize(0), F[k].clear(), k--;
        }
        ans += 2;
      }
    }
    printf("%d ", ans + dfs(1, k));
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int N = 1e5 + 7;
int n, k, w[N];
vector<int> vl[N], vr[N];
unordered_map<int, int> F[N];
inline int dfs(int l, int r) {
  if (F[r].count(l)) return F[r][l];
  int ans = 0;
  for (int i = 0; i < vl[l].size(); i++)
    if (l + vl[l][i] <= r) ans = max(ans, dfs(l + (vl[l][i] + 1) / 2, r) + 1);
  for (int i = 0; i < vr[r].size(); i++)
    if (r - vr[r][i] >= l) ans = max(ans, dfs(l, r - (vr[r][i] + 1) / 2) + 1);
  F[r][l] = ans;
  return ans;
}
int main() {
  cin >> n;
  int x, ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x), w[++k] = x;
    if (w[k] == w[k - 1]) vr[k].push_back(1), vl[k - 1].push_back(1);
    for (int j = 0; j < vr[k - 1].size(); j++) {
      int x = k - 2 - vr[k - 1][j];
      if (w[x] == w[k])
        vl[x].push_back(vr[k - 1][j] + 2), vr[k].push_back(vr[k - 1][j] + 2);
    }
    if (vr[k].size() == 1) {
      int p = k - (vr[k][0] + 1) / 2;
      bool fl = 0;
      for (int j = 0; j < vr[p].size(); j++)
        if (vr[p][j] == vr[k][0]) fl = 1;
      if (fl) {
        p = k - vr[k][0];
        while (k >= p) {
          for (int j = 0; j < vr[k].size(); j++)
            vl[k - vr[k][j]].resize(vl[k - vr[k][j]].size() - 1);
          vr[k].resize(0), F[k].clear(), k--;
        }
        ans += 2;
      }
    }
    printf("%d ", ans + dfs(1, k));
  }
}

```