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;
const int iinf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 3e5;
using edge = pair<int, int>;
int q;
long long ans;
map<edge, int> ti;
const int uN = N << 1;
int te[uN], sz[2][uN];
stack<int> st;
int find(int u) {
  while (u ^ te[u]) u = te[u];
  return u;
}
bool merge(int u, int v) {
  int first = find(u), second = find(v);
  if (first == second) return false;
  if (sz[0][first] + sz[1][first] > sz[0][second] + sz[1][second])
    swap(first, second);
  ans -=
      (1ll * sz[0][first] * sz[1][first] + 1ll * sz[0][second] * sz[1][second]);
  sz[0][second] += sz[0][first], sz[1][second] += sz[1][first];
  ans += 1ll * sz[0][second] * sz[1][second];
  te[first] = second, st.push(first);
  return true;
}
void rev() {
  int u = st.top();
  st.pop();
  ans -= 1ll * sz[0][te[u]] * sz[1][te[u]];
  sz[0][te[u]] -= sz[0][u], sz[1][te[u]] -= sz[1][u];
  ans += 1ll * sz[0][u] * sz[1][u] + 1ll * sz[0][te[u]] * sz[1][te[u]];
  te[u] = u;
}
vector<edge> ev[N << 2];
void adde(int first, int second, edge e, int k = 0, int l = 0, int r = q) {
  if (r <= first || second <= l) return;
  if (first <= l && r <= second) return ev[k].push_back(e), void();
  adde(first, second, e, k * 2 + 1, l, ((l + r) >> 1)),
      adde(first, second, e, k * 2 + 2, ((l + r) >> 1), r);
}
void divide(int k = 0, int l = 0, int r = q) {
  int mc = 0;
  for (edge e : ev[k])
    if (merge(e.first, e.second)) mc++;
  if (r - l == 1)
    cout << ans << " ";
  else
    divide(k * 2 + 1, l, ((l + r) >> 1)), divide(k * 2 + 2, ((l + r) >> 1), r);
  while (mc--) rev();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> q;
  iota(te, te + (N << 1), 0);
  fill(sz[0], sz[0] + N, 1);
  fill(sz[1] + N, sz[1] + (N << 1), 1);
  for (int i = (0), iE = (q); i < iE; i++) {
    edge e;
    cin >> e.first >> e.second, --e.first, --e.second, e.second += N;
    if (!ti.count(e) || !~ti[e])
      ti[e] = i;
    else
      adde(ti[e], i, e), ti[e] = -1;
  }
  for (auto t : ti)
    if (~t.second) adde(t.second, q, t.first);
  divide(), cout << '\n';
  return 0;
}