#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
vector<pair<int, int> > e;
set<pair<int, int> > f;
vector<int> adj[MAXN];
int n, m, k, deg[MAXN], num;
bool kicked[MAXN];
vector<int> ans;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    e.push_back(pair<int, int>(u, v));
    f.insert(pair<int, int>(u, v));
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  num = n;
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] < k) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int top = q.front();
    q.pop();
    if (kicked[top]) continue;
    kicked[top] = true;
    num--;
    for (int j : adj[top])
      if (f.count(pair<int, int>(top, j)) + f.count(pair<int, int>(j, top)) ==
          1) {
        deg[j]--;
        f.erase(pair<int, int>(j, top));
        f.erase(pair<int, int>(top, j));
        if (deg[j] < k) {
          q.push(j);
        }
      }
  }
  reverse(e.begin(), e.end());
  for (pair<int, int> i : e) {
    ans.push_back(num);
    if (f.count(i) == 0) continue;
    deg[i.first]--;
    deg[i.second]--;
    f.erase(i);
    queue<int> q;
    if (deg[i.first] < k && !kicked[i.first]) q.push(i.first);
    if (deg[i.second] < k && !kicked[i.second]) q.push(i.second);
    while (!q.empty()) {
      int top = q.front();
      q.pop();
      if (kicked[top]) continue;
      kicked[top] = true;
      num--;
      for (int j : adj[top])
        if (f.count(pair<int, int>(top, j)) + f.count(pair<int, int>(j, top)) ==
            1) {
          deg[j]--;
          f.erase(pair<int, int>(j, top));
          f.erase(pair<int, int>(top, j));
          if (deg[j] < k) {
            q.push(j);
          }
        }
    }
  }
  reverse(ans.begin(), ans.end());
  for (int i : ans) printf("%d\n", i);
}

### Prompt

Create a solution in cpp for 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 = 2e5 + 5;
vector<pair<int, int> > e;
set<pair<int, int> > f;
vector<int> adj[MAXN];
int n, m, k, deg[MAXN], num;
bool kicked[MAXN];
vector<int> ans;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    e.push_back(pair<int, int>(u, v));
    f.insert(pair<int, int>(u, v));
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  num = n;
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] < k) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int top = q.front();
    q.pop();
    if (kicked[top]) continue;
    kicked[top] = true;
    num--;
    for (int j : adj[top])
      if (f.count(pair<int, int>(top, j)) + f.count(pair<int, int>(j, top)) ==
          1) {
        deg[j]--;
        f.erase(pair<int, int>(j, top));
        f.erase(pair<int, int>(top, j));
        if (deg[j] < k) {
          q.push(j);
        }
      }
  }
  reverse(e.begin(), e.end());
  for (pair<int, int> i : e) {
    ans.push_back(num);
    if (f.count(i) == 0) continue;
    deg[i.first]--;
    deg[i.second]--;
    f.erase(i);
    queue<int> q;
    if (deg[i.first] < k && !kicked[i.first]) q.push(i.first);
    if (deg[i.second] < k && !kicked[i.second]) q.push(i.second);
    while (!q.empty()) {
      int top = q.front();
      q.pop();
      if (kicked[top]) continue;
      kicked[top] = true;
      num--;
      for (int j : adj[top])
        if (f.count(pair<int, int>(top, j)) + f.count(pair<int, int>(j, top)) ==
            1) {
          deg[j]--;
          f.erase(pair<int, int>(j, top));
          f.erase(pair<int, int>(top, j));
          if (deg[j] < k) {
            q.push(j);
          }
        }
    }
  }
  reverse(ans.begin(), ans.end());
  for (int i : ans) printf("%d\n", i);
}

```