#include <bits/stdc++.h>
using std::cin;
using std::cout;
using std::vector;
const int N = 100054;
int s[N];
vector<int> f[N];
namespace zig {
void push(int i) {
  f[i].clear();
  if (s[i - 1] == s[i]) f[i].emplace_back(1);
  for (int h : f[i - 1])
    if (s[i - 2 * h - 1] == s[i]) f[i].emplace_back(h + 1);
}
int check(int i) {
  if (f[i].empty()) return 0;
  int h = f[i].front();
  return std::binary_search(f[i - h].begin(), f[i - h].end(), h) ? h : 0;
}
}  // namespace zig
namespace misc {
int suf[N], preP[N], preC[N];
void push(int i) {
  int L;
  suf[i] = (f[i].empty() ? 0 : f[i].front()), preP[i] = preP[i - 1],
  preC[i] = preC[i - 1];
  if (!((L = i - preP[i]) & 1) &&
      std::binary_search(f[i].begin(), f[i].end(), L >>= 1))
    preP[i] += L, ++preC[i];
}
int query_suf(int i) {
  int j, ret = preC[i], L = preP[i];
  for (j = i; suf[j] && L < j - suf[j]; j -= suf[j]) ++ret;
  return ret;
}
}  // namespace misc
int main() {
  int i, h, n, x, Z = 0, top = 0;
  std::ios::sync_with_stdio(false), cin.tie(NULL);
  cin >> n, *s = -1;
  for (i = 0; i < n; ++i) {
    cin >> x, s[++top] = x, zig::push(top);
    if ((h = zig::check(top)))
      ++Z, top -= 2 * h;
    else
      misc::push(top);
    cout << 2 * Z + misc::query_suf(top) << (i == n - 1 ? '\n' : ' ');
  }
  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 std::cin;
using std::cout;
using std::vector;
const int N = 100054;
int s[N];
vector<int> f[N];
namespace zig {
void push(int i) {
  f[i].clear();
  if (s[i - 1] == s[i]) f[i].emplace_back(1);
  for (int h : f[i - 1])
    if (s[i - 2 * h - 1] == s[i]) f[i].emplace_back(h + 1);
}
int check(int i) {
  if (f[i].empty()) return 0;
  int h = f[i].front();
  return std::binary_search(f[i - h].begin(), f[i - h].end(), h) ? h : 0;
}
}  // namespace zig
namespace misc {
int suf[N], preP[N], preC[N];
void push(int i) {
  int L;
  suf[i] = (f[i].empty() ? 0 : f[i].front()), preP[i] = preP[i - 1],
  preC[i] = preC[i - 1];
  if (!((L = i - preP[i]) & 1) &&
      std::binary_search(f[i].begin(), f[i].end(), L >>= 1))
    preP[i] += L, ++preC[i];
}
int query_suf(int i) {
  int j, ret = preC[i], L = preP[i];
  for (j = i; suf[j] && L < j - suf[j]; j -= suf[j]) ++ret;
  return ret;
}
}  // namespace misc
int main() {
  int i, h, n, x, Z = 0, top = 0;
  std::ios::sync_with_stdio(false), cin.tie(NULL);
  cin >> n, *s = -1;
  for (i = 0; i < n; ++i) {
    cin >> x, s[++top] = x, zig::push(top);
    if ((h = zig::check(top)))
      ++Z, top -= 2 * h;
    else
      misc::push(top);
    cout << 2 * Z + misc::query_suf(top) << (i == n - 1 ? '\n' : ' ');
  }
  return 0;
}

```