#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)2e5 + 5;
int n, m, k;
vector<pair<int, int> > out[maxn];
int la[maxn];
int pos[maxn];
int deg[maxn];
int add[maxn];
bool dead[maxn];
set<pair<int, int>, greater<pair<int, int> > > s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  fill(la, la + n, -1);
  for (auto i = 0; i < m; ++i) {
    int v, u;
    cin >> v >> u;
    --v;
    --u;
    out[v].push_back(make_pair(i, u));
    out[u].push_back(make_pair(i, v));
    ++deg[v];
    ++deg[u];
    if (deg[v] == k) {
      pos[v] = out[v].size() - 1;
      la[v] = i;
    }
    if (deg[u] == k) {
      pos[u] = out[u].size() - 1;
      la[u] = i;
    }
  }
  for (auto i = 0; i < n; ++i)
    if (la[i] == -1) la[i] = m;
  for (auto i = 0; i < n; ++i) s.insert(make_pair(la[i], i));
  while (!s.empty()) {
    int v = s.begin()->second;
    dead[v] = true;
    s.erase(s.begin());
    for (auto e : out[v]) {
      int u = e.second;
      if (dead[u] || la[u] == la[v]) continue;
      if (la[u] < e.first) continue;
      s.erase(make_pair(la[u], u));
      ++pos[u];
      while (pos[u] != (int)out[u].size() && dead[out[u][pos[u]].second])
        ++pos[u];
      if (pos[u] == (int)out[u].size())
        la[u] = m;
      else
        la[u] = out[u][pos[u]].first;
      la[u] = min(la[u], la[v]);
      s.insert(make_pair(la[u], u));
    }
  }
  for (auto i = 0; i < n; ++i) add[la[i]]++;
  int curr = 0;
  for (auto i = 0; i < m; ++i) {
    curr += add[i];
    cout << curr << '\n';
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.

We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold: 

  * Either this person does not go on the trip, 
  * Or at least k of his friends also go on the trip. 



Note that the friendship is not transitive. That is, if a and b are friends and b and c are friends, it does not necessarily imply that a and c are friends.

For each day, find the maximum number of people that can go on the trip on that day.

Input

The first line contains three integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 2 ⋅ 10^5, 1 ≤ k < n) — the number of people, the number of days and the number of friends each person on the trip should have in the group.

The i-th (1 ≤ i ≤ m) of the next m lines contains two integers x and y (1≤ x, y≤ n, x≠ y), meaning that persons x and y become friends on the morning of day i. It is guaranteed that x and y were not friends before.

Output

Print exactly m lines, where the i-th of them (1≤ i≤ m) contains the maximum number of people that can go on the trip on the evening of the day i.

Examples

Input

4 4 2
2 3
1 2
1 3
1 4


Output

0
0
3
3


Input

5 8 2
2 1
4 2
5 4
5 2
4 3
5 1
4 1
3 2


Output

0
0
0
3
3
4
4
5


Input

5 7 2
1 5
3 2
2 5
3 4
1 2
5 3
1 3


Output

0
0
0
0
3
4
4

Note

In the first example, 

  * 1,2,3 can go on day 3 and 4. 



In the second example, 

  * 2,4,5 can go on day 4 and 5. 
  * 1,2,4,5 can go on day 6 and 7. 
  * 1,2,3,4,5 can go on day 8. 



In the third example, 

  * 1,2,5 can go on day 5. 
  * 1,2,3,5 can go on day 6 and 7. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)2e5 + 5;
int n, m, k;
vector<pair<int, int> > out[maxn];
int la[maxn];
int pos[maxn];
int deg[maxn];
int add[maxn];
bool dead[maxn];
set<pair<int, int>, greater<pair<int, int> > > s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  fill(la, la + n, -1);
  for (auto i = 0; i < m; ++i) {
    int v, u;
    cin >> v >> u;
    --v;
    --u;
    out[v].push_back(make_pair(i, u));
    out[u].push_back(make_pair(i, v));
    ++deg[v];
    ++deg[u];
    if (deg[v] == k) {
      pos[v] = out[v].size() - 1;
      la[v] = i;
    }
    if (deg[u] == k) {
      pos[u] = out[u].size() - 1;
      la[u] = i;
    }
  }
  for (auto i = 0; i < n; ++i)
    if (la[i] == -1) la[i] = m;
  for (auto i = 0; i < n; ++i) s.insert(make_pair(la[i], i));
  while (!s.empty()) {
    int v = s.begin()->second;
    dead[v] = true;
    s.erase(s.begin());
    for (auto e : out[v]) {
      int u = e.second;
      if (dead[u] || la[u] == la[v]) continue;
      if (la[u] < e.first) continue;
      s.erase(make_pair(la[u], u));
      ++pos[u];
      while (pos[u] != (int)out[u].size() && dead[out[u][pos[u]].second])
        ++pos[u];
      if (pos[u] == (int)out[u].size())
        la[u] = m;
      else
        la[u] = out[u][pos[u]].first;
      la[u] = min(la[u], la[v]);
      s.insert(make_pair(la[u], u));
    }
  }
  for (auto i = 0; i < n; ++i) add[la[i]]++;
  int curr = 0;
  for (auto i = 0; i < m; ++i) {
    curr += add[i];
    cout << curr << '\n';
  }
  return 0;
}

```