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 
#include <bits/stdc++.h>
using namespace std;
const int MAXQ = 3 * 100100, MAXV = 2 * 3 * 100100;
map<pair<int, int>, int> m;
int q, x[MAXQ], y[MAXQ], r[MAXQ], uf[MAXV], h[MAXV], cntx[MAXV], cnty[MAXV];
long long ans[MAXQ], curans = 0;
stack<int> s;
void undo() {
  if (s.top() == 0) return void(s.pop());
  curans -= 1LL * cntx[uf[s.top()]] * cnty[uf[s.top()]];
  cntx[uf[s.top()]] -= cntx[s.top()], cnty[uf[s.top()]] -= cnty[s.top()];
  curans += 1LL * cntx[uf[s.top()]] * cnty[uf[s.top()]];
  curans += 1LL * cntx[s.top()] * cnty[s.top()];
  uf[s.top()] = 0;
  s.pop();
}
inline int find(int x) {
  while (uf[x]) x = uf[x];
  return x;
}
void join(int u, int v) {
  if ((u = find(u)) == (v = find(v))) return void(s.push(0));
  if (h[u] < h[v]) swap(u, v);
  s.push(v);
  curans -= 1LL * cntx[u] * cnty[u] + 1LL * cntx[v] * cnty[v];
  uf[v] = u, h[u] += (h[u] == h[v]), cntx[u] += cntx[v], cnty[u] += cnty[v];
  curans += 1LL * cntx[u] * cnty[u];
}
struct no {
  int beg, end, mid;
  no *l, *r;
  vector<pair<int, int>> e;
  no(int beg, int end) : beg(beg), end(end), mid((beg + end) >> 1) {
    if (beg != end) l = new no(beg, mid), r = new no(mid + 1, end);
  }
  ~no() {
    if (beg != end) delete l, delete r;
  }
  void upd(int a, int b, int u, int v) {
    if (b < beg || end < a) return;
    if (a <= beg && end <= b) return void(e.push_back(make_pair(u, v)));
    l->upd(a, b, u, v), r->upd(a, b, u, v);
  }
  void dfs() {
    for (const auto& [u, v] : e) join(u, v);
    if (beg != end)
      l->dfs(), r->dfs();
    else
      ans[beg] = curans;
    for (const auto& [u, v] : e) undo();
  }
};
int main() {
  scanf("%d", &q);
  no seg(1, q);
  for (int i = 1; i < MAXV; ++i)
    cntx[i] = (i <= (MAXV >> 1)), cnty[i] = (i > (MAXV >> 1));
  for (int i = 1; i <= q; ++i) {
    scanf("%d %d", &x[i], &y[i]);
    int& aux = m[make_pair(x[i], y[i])];
    if (aux)
      r[aux] = i - 1, r[i] = -1, aux = 0;
    else
      aux = i;
  }
  for (int i = 1; i <= q; ++i)
    seg.upd(i, (r[i] ? r[i] : q), x[i], y[i] + (MAXV >> 1));
  seg.dfs();
  for (int i = 1; i <= q; ++i) printf("%lld ", ans[i]);
}
