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>
const int md = 1e9 + 7;
const long long inf = 1e18;
const int OO = 0;
const int OOO = 1;
using namespace std;
struct info {
  int from, to, ra;
  info() { to = from = ra = -1; }
  info(int f, int t, int r) {
    from = f;
    to = t;
    ra = r;
  }
} org[300003];
long long operator*(pair<int, int> &aa, pair<int, int> &bb) {
  return (long long)aa.first * bb.second + (long long)aa.second * bb.first;
}
int p[600000], r[600000] = {};
pair<int, int> cnt[600000];
struct dsu {
  int n;
  int nxt = 0;
  dsu() {}
  dsu(int sz) {
    n = sz;
    nxt = 0;
    for (int i = 0; i < n; i++) {
      p[i] = i;
      cnt[i] = (i < n / 2 ? make_pair(1, 0) : make_pair(0, 1));
    }
  }
  int find(int x) {
    while (x != p[x]) x = p[x];
    return x;
  }
  long long unite(int x, int y) {
    x--, y--;
    x = find(x), y = find(y);
    if (x == y) {
      org[nxt++] = info();
      return 0;
    }
    long long rtn = cnt[x] * cnt[y];
    if (r[y] > r[x]) {
      p[x] = y;
      org[nxt++] = (info(x, y, r[y]));
      cnt[y].first += cnt[x].first;
      cnt[y].second += cnt[x].second;
    } else {
      p[y] = x;
      org[nxt++] = (info(y, x, r[x]));
      if (r[x] == r[y]) r[x]++;
      cnt[x].first += cnt[y].first;
      cnt[x].second += cnt[y].second;
    }
    return rtn;
  }
  long long undo() {
    if (nxt == 0) return 0;
    info i = org[--nxt];
    if (i.from == -1) return 0;
    p[i.from] = i.from;
    r[i.to] = i.ra;
    cnt[i.to].first -= cnt[i.from].first;
    cnt[i.to].second -= cnt[i.from].second;
    return cnt[i.from] * cnt[i.to];
  }
};
long long H(pair<int, int> aa) { return 300003LL * aa.first + aa.second; }
pair<int, int> H(long long aa) { return {aa / 300003, aa % 300003}; }
const int nn = 1 << 19;
int n, pr = 0;
long long ans = 0;
dsu D(600000);
map<long long, int> M;
vector<pair<int, int>> tree[2 * nn];
void add(int st, int en, const pair<int, int> &val, int node, int nl, int nr) {
  if (en < nl || nr < st) return;
  if (st <= nl && nr <= en) {
    if (OO) {
      cout << "added in " << node << " " << nl << " " << nr << '\n';
    }
    tree[node].push_back(val);
    return;
  }
  int mid = (nl + nr) / 2;
  add(st, en, val, 2 * node + 1, nl, mid);
  add(st, en, val, 2 * node + 2, mid + 1, nr);
}
void solve(int node, int nl, int nr) {
  if (nl >= n) return;
  for (auto &i : tree[node]) {
    ans += D.unite(i.first, 300000 + i.second);
  }
  if (2 * node + 2 < 2 * nn) {
    int mid = (nl + nr) / 2;
    solve(2 * node + 1, nl, mid);
    solve(2 * node + 2, mid + 1, nr);
  } else {
    cout << ans << " ";
    pr++;
    if (pr == n) {
      cout << '\n';
      exit(0);
    }
  }
  for (int i = 0; i < tree[node].size(); i++) ans -= D.undo();
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    pair<int, int> pa = {x, y};
    long long h = H(pa);
    auto it = M.find(h);
    if (it != M.end()) {
      if (OO) {
        cout << "adding " << it->second << " " << i - 1 << " the pair " << x
             << " " << y << '\n';
      }
      add(it->second, i - 1, pa, 0, 0, nn - 1);
      M.erase(it);
    } else {
      M[h] = i;
    }
  }
  for (auto &i : M) {
    pair<int, int> pa = H(i.first);
    add(i.second, n - 1, pa, 0, 0, nn - 1);
  }
  solve(0, 0, nn - 1);
}