#include <bits/stdc++.h>
using namespace std;
int w[101000], n, D[101000], S, Right[101000];
vector<int> G[101000];
vector<int> V[101000];
int main() {
  int i, Q;
  scanf("%d", &Q);
  while (Q--) {
    n++;
    scanf("%d", &w[n]);
    if (n > 1 && w[n] == w[n - 1]) V[n].push_back(n - 1);
    for (auto &t : V[n - 1]) {
      if (w[n] != w[t - 1]) continue;
      V[n].push_back(t - 1);
    }
    for (auto &t : V[n]) {
      if (!Right[t - 1]) {
        Right[t - 1] = n;
        G[n].push_back(t - 1);
      }
    }
    while (1) {
      int r = -1;
      for (auto &t : V[n]) {
        int l = (n - t + 1) / 2;
        int x = n - l, ck = 0;
        for (auto &y : V[x]) {
          if (x - y == n - t) {
            ck = 1;
            break;
          }
        }
        if (ck) {
          r = l * 2;
          break;
        }
      }
      if (r == -1) break;
      for (i = 0; i < r; i++) {
        for (auto &t : G[n]) Right[t] = 0;
        G[n].clear();
        V[n].clear();
        n--;
      }
      S += 2;
    }
    int rr = n, c = 0;
    while (!V[rr].empty()) {
      rr -= (rr - V[rr][0] + 1) / 2;
      c++;
    }
    int l = 0;
    while (Right[l] && Right[l] <= rr) {
      l += (Right[l] - l) / 2;
      c++;
    }
    printf("%d ", S + c);
  }
}

### Prompt

Develop a solution in CPP 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;
int w[101000], n, D[101000], S, Right[101000];
vector<int> G[101000];
vector<int> V[101000];
int main() {
  int i, Q;
  scanf("%d", &Q);
  while (Q--) {
    n++;
    scanf("%d", &w[n]);
    if (n > 1 && w[n] == w[n - 1]) V[n].push_back(n - 1);
    for (auto &t : V[n - 1]) {
      if (w[n] != w[t - 1]) continue;
      V[n].push_back(t - 1);
    }
    for (auto &t : V[n]) {
      if (!Right[t - 1]) {
        Right[t - 1] = n;
        G[n].push_back(t - 1);
      }
    }
    while (1) {
      int r = -1;
      for (auto &t : V[n]) {
        int l = (n - t + 1) / 2;
        int x = n - l, ck = 0;
        for (auto &y : V[x]) {
          if (x - y == n - t) {
            ck = 1;
            break;
          }
        }
        if (ck) {
          r = l * 2;
          break;
        }
      }
      if (r == -1) break;
      for (i = 0; i < r; i++) {
        for (auto &t : G[n]) Right[t] = 0;
        G[n].clear();
        V[n].clear();
        n--;
      }
      S += 2;
    }
    int rr = n, c = 0;
    while (!V[rr].empty()) {
      rr -= (rr - V[rr][0] + 1) / 2;
      c++;
    }
    int l = 0;
    while (Right[l] && Right[l] <= rr) {
      l += (Right[l] - l) / 2;
      c++;
    }
    printf("%d ", S + c);
  }
}

```