#include <bits/stdc++.h>
using namespace std;
const int maxN = 2e5 + 11;
typedef int i_N[maxN];
int N, M, K;
set<int> adj[maxN];
i_N deg, mark;
int ans;
pair<int, int> edge[maxN];
void dfs(int u) {
  if (mark[u]++) return;
  ans--;
  for (set<int>::iterator it = adj[u].begin(); it != adj[u].end(); it++) {
    int v = *it;
    deg[v]--;
    adj[v].erase(u);
    if (deg[v] < K) dfs(v);
  }
  adj[u].clear();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> M >> K;
  for (int i = 1; i <= M; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].insert(v);
    adj[v].insert(u);
    deg[u]++;
    deg[v]++;
    edge[i] = pair<int, int>(u, v);
  }
  ans = N;
  for (int i = 1; i <= N; i++)
    if (deg[i] < K) dfs(i);
  vector<int> v;
  for (int i = M; i; i--) {
    v.push_back(ans);
    int u = edge[i].first, v = edge[i].second;
    if (adj[u].find(v) != adj[u].end()) {
      deg[u]--;
      deg[v]--;
      adj[u].erase(v);
      adj[v].erase(u);
    }
    if (deg[u] < K) dfs(u);
    if (deg[v] < K) dfs(v);
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < M; i++) cout << v[i] << '\n';
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 2e5 + 11;
typedef int i_N[maxN];
int N, M, K;
set<int> adj[maxN];
i_N deg, mark;
int ans;
pair<int, int> edge[maxN];
void dfs(int u) {
  if (mark[u]++) return;
  ans--;
  for (set<int>::iterator it = adj[u].begin(); it != adj[u].end(); it++) {
    int v = *it;
    deg[v]--;
    adj[v].erase(u);
    if (deg[v] < K) dfs(v);
  }
  adj[u].clear();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> M >> K;
  for (int i = 1; i <= M; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].insert(v);
    adj[v].insert(u);
    deg[u]++;
    deg[v]++;
    edge[i] = pair<int, int>(u, v);
  }
  ans = N;
  for (int i = 1; i <= N; i++)
    if (deg[i] < K) dfs(i);
  vector<int> v;
  for (int i = M; i; i--) {
    v.push_back(ans);
    int u = edge[i].first, v = edge[i].second;
    if (adj[u].find(v) != adj[u].end()) {
      deg[u]--;
      deg[v]--;
      adj[u].erase(v);
      adj[v].erase(u);
    }
    if (deg[u] < K) dfs(u);
    if (deg[v] < K) dfs(v);
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < M; i++) cout << v[i] << '\n';
}

```