#include <bits/stdc++.h>
using namespace std;
template <class u>
pair<u, u> ordered_pair(const u& a, const u& b) {
  return a < b ? make_pair(a, b) : make_pair(b, a);
}
int n, m, k;
set<int> gr[201010];
vector<pair<int, int>> edge_list;
set<pair<int, int>> removed_edges;
struct cmp_deg {
  bool operator()(int u, int v) {
    return make_pair(((int)gr[u].size()), u) <
           make_pair(((int)gr[v].size()), v);
  }
};
set<int, cmp_deg> node_set;
void remove_edge(int u, int v) {
  node_set.erase(u);
  node_set.erase(v);
  gr[u].erase(v);
  gr[v].erase(u);
  node_set.insert(u);
  node_set.insert(v);
  removed_edges.insert(ordered_pair(u, v));
}
void remove_node(int u) {
  while (((int)gr[u].size())) remove_edge(u, *gr[u].begin());
  node_set.erase(u);
}
void adjust() {
  while (((int)node_set.size()) and ((int)gr[*node_set.begin()].size()) < k) {
    remove_node(*node_set.begin());
  }
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = -1; ++i < m;) {
    int u, v;
    cin >> u >> v;
    gr[u].insert(v);
    gr[v].insert(u);
    edge_list.push_back(ordered_pair(u, v));
  }
  for (int i = 0; i++ < n;) node_set.insert(i);
  vector<int> ans(m);
  for (int i = m; i--;) {
    adjust();
    ans[i] = ((int)node_set.size());
    if (!removed_edges.count(edge_list[i])) {
      remove_edge(edge_list[i].first, edge_list[i].second);
    }
  }
  for (auto i : ans) cout << i << '\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;
template <class u>
pair<u, u> ordered_pair(const u& a, const u& b) {
  return a < b ? make_pair(a, b) : make_pair(b, a);
}
int n, m, k;
set<int> gr[201010];
vector<pair<int, int>> edge_list;
set<pair<int, int>> removed_edges;
struct cmp_deg {
  bool operator()(int u, int v) {
    return make_pair(((int)gr[u].size()), u) <
           make_pair(((int)gr[v].size()), v);
  }
};
set<int, cmp_deg> node_set;
void remove_edge(int u, int v) {
  node_set.erase(u);
  node_set.erase(v);
  gr[u].erase(v);
  gr[v].erase(u);
  node_set.insert(u);
  node_set.insert(v);
  removed_edges.insert(ordered_pair(u, v));
}
void remove_node(int u) {
  while (((int)gr[u].size())) remove_edge(u, *gr[u].begin());
  node_set.erase(u);
}
void adjust() {
  while (((int)node_set.size()) and ((int)gr[*node_set.begin()].size()) < k) {
    remove_node(*node_set.begin());
  }
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = -1; ++i < m;) {
    int u, v;
    cin >> u >> v;
    gr[u].insert(v);
    gr[v].insert(u);
    edge_list.push_back(ordered_pair(u, v));
  }
  for (int i = 0; i++ < n;) node_set.insert(i);
  vector<int> ans(m);
  for (int i = m; i--;) {
    adjust();
    ans[i] = ((int)node_set.size());
    if (!removed_edges.count(edge_list[i])) {
      remove_edge(edge_list[i].first, edge_list[i].second);
    }
  }
  for (auto i : ans) cout << i << '\n';
  return 0;
}

```