#include <bits/stdc++.h>
using namespace std;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
int n;
int a[MAXN];
int ans[MAXN];
int b[MAXN], nb;
vector<int> palsuffixes[MAXN];
int shortestpal[MAXN];
vector<int> shortestpalends[MAXN];
void solve() {
  nb = 0;
  int sofar = 0;
  for (int i = (0); i < (n); ++i) {
    int x = a[i];
    if (nb - 1 >= 0) {
      for (int j = (0); j < (((int)(palsuffixes[nb - 1]).size())); ++j) {
        int plen = palsuffixes[nb - 1][j];
        if (nb - plen - 1 >= 0 && b[nb - plen - 1] == x)
          palsuffixes[nb].push_back(plen + 2);
      }
      if (b[nb - 1] == x) palsuffixes[nb].push_back(2);
    }
    b[nb++] = x;
    shortestpal[nb - 1] = -1;
    for (int j = (0); j < (((int)(palsuffixes[nb - 1]).size())); ++j) {
      int len = palsuffixes[nb - 1][j];
      if (shortestpal[nb - len] == -1)
        shortestpal[nb - len] = len,
                         shortestpalends[nb - 1].push_back(nb - len);
    }
    if (((int)(palsuffixes[nb - 1]).size()) >= 1) {
      int mnlen = palsuffixes[nb - 1].back();
      int pidx = nb - mnlen / 2 - 1;
      bool found = false;
      for (int j = (0); j < (((int)(palsuffixes[pidx]).size())); ++j)
        if (palsuffixes[pidx][j] == mnlen) found = true;
      if (found) {
        sofar += 2;
        for (int j = (0); j < (mnlen); ++j) {
          --nb;
          palsuffixes[nb].clear();
          for (int k = (0); k < (((int)(shortestpalends[nb]).size())); ++k) {
            int idx = shortestpalends[nb][k];
            assert(shortestpal[idx] == nb - idx + 1);
            shortestpal[idx] = -1;
          }
          shortestpalends[nb].clear();
        }
      }
    }
    ans[i] = sofar;
    int prefrem = 0;
    {
      int at = 0;
      while (shortestpal[at] != -1)
        ++ans[i], prefrem += shortestpal[at] / 2, at += shortestpal[at] / 2;
    }
    {
      int at = nb - 1;
      while (((int)(palsuffixes[at]).size()) >= 1) {
        int len = palsuffixes[at].back();
        if (at + 1 - prefrem < len) break;
        ++ans[i], at -= len / 2;
      }
    }
  }
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  solve();
  for (int i = (0); i < (n); ++i) {
    if (i != 0) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
}
int main() {
  run();
  return 0;
}

### Prompt

Create a solution in cpp for 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;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
int n;
int a[MAXN];
int ans[MAXN];
int b[MAXN], nb;
vector<int> palsuffixes[MAXN];
int shortestpal[MAXN];
vector<int> shortestpalends[MAXN];
void solve() {
  nb = 0;
  int sofar = 0;
  for (int i = (0); i < (n); ++i) {
    int x = a[i];
    if (nb - 1 >= 0) {
      for (int j = (0); j < (((int)(palsuffixes[nb - 1]).size())); ++j) {
        int plen = palsuffixes[nb - 1][j];
        if (nb - plen - 1 >= 0 && b[nb - plen - 1] == x)
          palsuffixes[nb].push_back(plen + 2);
      }
      if (b[nb - 1] == x) palsuffixes[nb].push_back(2);
    }
    b[nb++] = x;
    shortestpal[nb - 1] = -1;
    for (int j = (0); j < (((int)(palsuffixes[nb - 1]).size())); ++j) {
      int len = palsuffixes[nb - 1][j];
      if (shortestpal[nb - len] == -1)
        shortestpal[nb - len] = len,
                         shortestpalends[nb - 1].push_back(nb - len);
    }
    if (((int)(palsuffixes[nb - 1]).size()) >= 1) {
      int mnlen = palsuffixes[nb - 1].back();
      int pidx = nb - mnlen / 2 - 1;
      bool found = false;
      for (int j = (0); j < (((int)(palsuffixes[pidx]).size())); ++j)
        if (palsuffixes[pidx][j] == mnlen) found = true;
      if (found) {
        sofar += 2;
        for (int j = (0); j < (mnlen); ++j) {
          --nb;
          palsuffixes[nb].clear();
          for (int k = (0); k < (((int)(shortestpalends[nb]).size())); ++k) {
            int idx = shortestpalends[nb][k];
            assert(shortestpal[idx] == nb - idx + 1);
            shortestpal[idx] = -1;
          }
          shortestpalends[nb].clear();
        }
      }
    }
    ans[i] = sofar;
    int prefrem = 0;
    {
      int at = 0;
      while (shortestpal[at] != -1)
        ++ans[i], prefrem += shortestpal[at] / 2, at += shortestpal[at] / 2;
    }
    {
      int at = nb - 1;
      while (((int)(palsuffixes[at]).size()) >= 1) {
        int len = palsuffixes[at].back();
        if (at + 1 - prefrem < len) break;
        ++ans[i], at -= len / 2;
      }
    }
  }
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  solve();
  for (int i = (0); i < (n); ++i) {
    if (i != 0) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
}
int main() {
  run();
  return 0;
}

```