Problem: For a given set of two-dimensional points S, let's denote its extension E(S) as the result of the following algorithm:

Create another set of two-dimensional points R, which is initially equal to S. Then, while there exist four numbers x_1, y_1, x_2 and y_2 such that (x_1, y_1) ∈ R, (x_1, y_2) ∈ R, (x_2, y_1) ∈ R and (x_2, y_2) ∉ R, add (x_2, y_2) to R. When it is impossible to find such four integers, let R be the result of the algorithm.

Now for the problem itself. You are given a set of two-dimensional points S, which is initially empty. You have to process two types of queries: add some point to S, or remove some point from it. After each query you have to compute the size of E(S).

Input

The first line contains one integer q (1 ≤ q ≤ 3 ⋅ 10^5) — the number of queries.

Then q lines follow, each containing two integers x_i, y_i (1 ≤ x_i, y_i ≤ 3 ⋅ 10^5), denoting i-th query as follows: if (x_i, y_i) ∈ S, erase it from S, otherwise insert (x_i, y_i) into S.

Output

Print q integers. i-th integer should be equal to the size of E(S) after processing first i queries.

Example

Input


7
1 1
1 2
2 1
2 2
1 2
1 3
2 1


Output


1 2 4 4 4 6 3 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MX = 3e5 + 1;
int par[MX * 2], szx[MX * 2], szy[MX * 2], n;
vector<pair<int, int>> stk, st[MX * 4];
ll res, ans[MX];
int find(int a) {
  if (a == par[a]) return a;
  return find(par[a]);
}
bool unite(int a, int b) {
  a = find(a);
  b = find(b);
  if (a == b) return false;
  if (szx[a] + szy[a] > szx[b] + szy[b]) {
    swap(a, b);
  }
  par[a] = b;
  res -= (ll)szx[a] * szy[a] + (ll)szx[b] * szy[b];
  szx[b] += szx[a];
  szy[b] += szy[a];
  res += (ll)szx[b] * szy[b];
  stk.emplace_back(a, b);
  return true;
}
void rollback() {
  auto [a, b] = stk.back();
  stk.pop_back();
  res -= (ll)szx[b] * szy[b];
  szx[b] -= szx[a];
  szy[b] -= szy[a];
  res += (ll)szx[b] * szy[b] + (ll)szx[a] * szy[a];
  par[a] = a;
}
void upd(int a, int b, pair<int, int> val, int v = 1, int l = 0,
         int r = n - 1) {
  if (b < l || r < a) return;
  if (a <= l && r <= b) {
    st[v].push_back(val);
  } else {
    int m = (l + r) / 2;
    upd(a, b, val, v * 2, l, m);
    upd(a, b, val, v * 2 + 1, m + 1, r);
  }
}
void dfs(int v = 1, int l = 0, int r = n - 1) {
  int cnt = 0;
  for (auto [x, y] : st[v]) {
    cnt += unite(x, MX + y);
  }
  if (l == r) {
    ans[l] = res;
  } else {
    int m = (l + r) / 2;
    dfs(v * 2, l, m);
    dfs(v * 2 + 1, m + 1, r);
  }
  while (cnt--) {
    rollback();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  map<pair<int, int>, int> mp;
  for (int i = 0; i < n; ++i) {
    pair<int, int> q;
    cin >> q.first >> q.second;
    if (mp.count(q)) {
      upd(mp[q], i - 1, q);
      mp.erase(q);
    } else {
      mp[q] = i;
    }
  }
  for (auto [q, t] : mp) {
    upd(t, n - 1, q);
  }
  iota(par, par + 2 * MX, 0);
  fill(szx + 1, szx + MX, 1);
  fill(szy + 1 + MX, szy + 2 * MX, 1);
  dfs();
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
}