#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int n, m, k, deg[N], ans[N], s[N], t[N];
map<int, bool> adj[N];
vector<int> g[N];
bool alive[N];
void solve() {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  memset(alive, 1, sizeof(alive));
  for (int u = 1; u <= n; u++) q.push(make_pair(deg[u], u));
  while (q.size()) {
    pair<int, int> now = q.top();
    if (now.first >= k) break;
    q.pop();
    if (!alive[now.second]) continue;
    alive[now.second] = 0;
    for (int &v : g[now.second]) {
      deg[v]--;
      if (alive[v]) q.push(make_pair(deg[v], v));
    }
  }
  int res = 0;
  for (int u = 1; u <= n; u++) res += alive[u];
  ans[m] = res;
  for (int i = m; i >= 1; i--) {
    if (alive[s[i]] && alive[t[i]]) {
      deg[s[i]]--;
      deg[t[i]]--;
      adj[s[i]][t[i]] = 0;
      adj[t[i]][s[i]] = 0;
      q.push(make_pair(deg[s[i]], s[i]));
      q.push(make_pair(deg[t[i]], t[i]));
      pair<int, int> now = q.top();
      while (q.size()) {
        pair<int, int> now = q.top();
        if (now.first >= k) break;
        q.pop();
        if (!alive[now.second]) continue;
        --res;
        alive[now.second] = 0;
        for (int &v : g[now.second]) {
          if (!adj[now.second][v]) continue;
          --deg[v];
          if (alive[v]) q.push(make_pair(deg[v], v));
        }
      }
    }
    ans[i - 1] = res;
  }
  for (int i = 1; i <= m; i++) printf("%d\n ", ans[i]);
}
int main() {
  scanf("%d %d %d ", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d ", &s[i], &t[i]);
    deg[s[i]]++;
    deg[t[i]]++;
    adj[s[i]][t[i]] = adj[t[i]][s[i]] = 1;
    g[s[i]].push_back(t[i]);
    g[t[i]].push_back(s[i]);
  }
  solve();
}

### 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 N = 2e5 + 7;
int n, m, k, deg[N], ans[N], s[N], t[N];
map<int, bool> adj[N];
vector<int> g[N];
bool alive[N];
void solve() {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  memset(alive, 1, sizeof(alive));
  for (int u = 1; u <= n; u++) q.push(make_pair(deg[u], u));
  while (q.size()) {
    pair<int, int> now = q.top();
    if (now.first >= k) break;
    q.pop();
    if (!alive[now.second]) continue;
    alive[now.second] = 0;
    for (int &v : g[now.second]) {
      deg[v]--;
      if (alive[v]) q.push(make_pair(deg[v], v));
    }
  }
  int res = 0;
  for (int u = 1; u <= n; u++) res += alive[u];
  ans[m] = res;
  for (int i = m; i >= 1; i--) {
    if (alive[s[i]] && alive[t[i]]) {
      deg[s[i]]--;
      deg[t[i]]--;
      adj[s[i]][t[i]] = 0;
      adj[t[i]][s[i]] = 0;
      q.push(make_pair(deg[s[i]], s[i]));
      q.push(make_pair(deg[t[i]], t[i]));
      pair<int, int> now = q.top();
      while (q.size()) {
        pair<int, int> now = q.top();
        if (now.first >= k) break;
        q.pop();
        if (!alive[now.second]) continue;
        --res;
        alive[now.second] = 0;
        for (int &v : g[now.second]) {
          if (!adj[now.second][v]) continue;
          --deg[v];
          if (alive[v]) q.push(make_pair(deg[v], v));
        }
      }
    }
    ans[i - 1] = res;
  }
  for (int i = 1; i <= m; i++) printf("%d\n ", ans[i]);
}
int main() {
  scanf("%d %d %d ", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d ", &s[i], &t[i]);
    deg[s[i]]++;
    deg[t[i]]++;
    adj[s[i]][t[i]] = adj[t[i]][s[i]] = 1;
    g[s[i]].push_back(t[i]);
    g[t[i]].push_back(s[i]);
  }
  solve();
}

```