#include <bits/stdc++.h>
using namespace std;
template <typename T, typename Pr = less<T>>
using pq = priority_queue<T, vector<T>, Pr>;
using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;
set<int> adj[200005];
bool removed[200005];
int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  vector<ii> edges(m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &edges[i].first, &edges[i].second);
    adj[edges[i].first].insert(edges[i].second);
    adj[edges[i].second].insert(edges[i].first);
  }
  set<ii> nodes;
  for (int i = 1; i <= n; i++) nodes.emplace(adj[i].size(), i);
  reverse((edges).begin(), (edges).end());
  vector<int> ans(m);
  for (int i = 0; i < m; i++) {
    while (!nodes.empty() && nodes.begin()->first < k) {
      auto x = nodes.begin()->second;
      removed[x] = true;
      nodes.erase(nodes.begin());
      for (auto& e : adj[x]) {
        nodes.erase(ii(adj[e].size(), e));
        adj[e].erase(x);
        nodes.emplace(adj[e].size(), e);
      }
    }
    ans[m - 1 - i] = nodes.size();
    auto& [x, y] = edges[i];
    if (removed[x] || removed[y]) continue;
    nodes.erase(ii(adj[x].size(), x));
    nodes.erase(ii(adj[y].size(), y));
    adj[x].erase(y);
    adj[y].erase(x);
    nodes.emplace(adj[x].size(), x);
    nodes.emplace(adj[y].size(), y);
  }
  for (auto& ai : ans) printf("%d\n", ai);
  return 0;
}

### Prompt

Please formulate 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;
template <typename T, typename Pr = less<T>>
using pq = priority_queue<T, vector<T>, Pr>;
using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;
set<int> adj[200005];
bool removed[200005];
int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  vector<ii> edges(m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &edges[i].first, &edges[i].second);
    adj[edges[i].first].insert(edges[i].second);
    adj[edges[i].second].insert(edges[i].first);
  }
  set<ii> nodes;
  for (int i = 1; i <= n; i++) nodes.emplace(adj[i].size(), i);
  reverse((edges).begin(), (edges).end());
  vector<int> ans(m);
  for (int i = 0; i < m; i++) {
    while (!nodes.empty() && nodes.begin()->first < k) {
      auto x = nodes.begin()->second;
      removed[x] = true;
      nodes.erase(nodes.begin());
      for (auto& e : adj[x]) {
        nodes.erase(ii(adj[e].size(), e));
        adj[e].erase(x);
        nodes.emplace(adj[e].size(), e);
      }
    }
    ans[m - 1 - i] = nodes.size();
    auto& [x, y] = edges[i];
    if (removed[x] || removed[y]) continue;
    nodes.erase(ii(adj[x].size(), x));
    nodes.erase(ii(adj[y].size(), y));
    adj[x].erase(y);
    adj[y].erase(x);
    nodes.emplace(adj[x].size(), x);
    nodes.emplace(adj[y].size(), y);
  }
  for (auto& ai : ans) printf("%d\n", ai);
  return 0;
}

```