#include <bits/stdc++.h>
using namespace std;
int n, m, k;
int a[200005];
int b[200005];
vector<int> adj[200005];
bool removed[200005];
int nc[200005];
int ans[200005];
set<pair<int, int>> remedges;
int main() {
  ios::sync_with_stdio(false);
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a[i], &b[i]);
  }
  for (int i = 0; i < m; i++) {
    adj[a[i]].push_back(b[i]);
    adj[b[i]].push_back(a[i]);
    nc[a[i]]++;
    nc[b[i]]++;
  }
  int cans = n;
  for (int i = 1; i <= n; i++) {
    if (!removed[i]) {
      if (nc[i] < k) {
        removed[i] = true;
        cans--;
        queue<int> q;
        q.push(i);
        while (!q.empty()) {
          int cur = q.front();
          q.pop();
          for (int j = 0; j < adj[cur].size(); j++) {
            int x = adj[cur][j];
            if (!removed[x]) {
              nc[x]--;
              if (nc[x] < k) {
                removed[x] = true;
                cans--;
                q.push(x);
              }
            }
          }
        }
      }
    }
  }
  ans[m - 1] = cans;
  for (int cm = m - 1; cm >= 0; cm--) {
    int ca = a[cm];
    int cb = b[cm];
    if (!removed[ca] && !removed[cb]) {
      nc[ca]--;
      nc[cb]--;
      remedges.insert(make_pair(ca, cb));
      remedges.insert(make_pair(cb, ca));
      if (nc[ca] < k) {
        removed[ca] = true;
        cans--;
        queue<int> q;
        q.push(ca);
        while (!q.empty()) {
          int cur = q.front();
          q.pop();
          for (int j = 0; j < adj[cur].size(); j++) {
            int x = adj[cur][j];
            if (remedges.find(make_pair(cur, x)) != remedges.end()) {
              continue;
            }
            if (!removed[x]) {
              nc[x]--;
              if (nc[x] < k) {
                removed[x] = true;
                cans--;
                q.push(x);
              }
            }
          }
        }
      }
      if (nc[cb] < k && !removed[cb]) {
        removed[cb] = true;
        cans--;
        queue<int> q;
        q.push(cb);
        while (!q.empty()) {
          int cur = q.front();
          q.pop();
          for (int j = 0; j < adj[cur].size(); j++) {
            int x = adj[cur][j];
            if (remedges.find(make_pair(cur, x)) != remedges.end()) {
              continue;
            }
            if (!removed[x]) {
              nc[x]--;
              if (nc[x] < k) {
                removed[x] = true;
                cans--;
                q.push(x);
              }
            }
          }
        }
      }
    }
    if (cm > 0) {
      ans[cm - 1] = cans;
    }
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
}

### 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;
int n, m, k;
int a[200005];
int b[200005];
vector<int> adj[200005];
bool removed[200005];
int nc[200005];
int ans[200005];
set<pair<int, int>> remedges;
int main() {
  ios::sync_with_stdio(false);
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a[i], &b[i]);
  }
  for (int i = 0; i < m; i++) {
    adj[a[i]].push_back(b[i]);
    adj[b[i]].push_back(a[i]);
    nc[a[i]]++;
    nc[b[i]]++;
  }
  int cans = n;
  for (int i = 1; i <= n; i++) {
    if (!removed[i]) {
      if (nc[i] < k) {
        removed[i] = true;
        cans--;
        queue<int> q;
        q.push(i);
        while (!q.empty()) {
          int cur = q.front();
          q.pop();
          for (int j = 0; j < adj[cur].size(); j++) {
            int x = adj[cur][j];
            if (!removed[x]) {
              nc[x]--;
              if (nc[x] < k) {
                removed[x] = true;
                cans--;
                q.push(x);
              }
            }
          }
        }
      }
    }
  }
  ans[m - 1] = cans;
  for (int cm = m - 1; cm >= 0; cm--) {
    int ca = a[cm];
    int cb = b[cm];
    if (!removed[ca] && !removed[cb]) {
      nc[ca]--;
      nc[cb]--;
      remedges.insert(make_pair(ca, cb));
      remedges.insert(make_pair(cb, ca));
      if (nc[ca] < k) {
        removed[ca] = true;
        cans--;
        queue<int> q;
        q.push(ca);
        while (!q.empty()) {
          int cur = q.front();
          q.pop();
          for (int j = 0; j < adj[cur].size(); j++) {
            int x = adj[cur][j];
            if (remedges.find(make_pair(cur, x)) != remedges.end()) {
              continue;
            }
            if (!removed[x]) {
              nc[x]--;
              if (nc[x] < k) {
                removed[x] = true;
                cans--;
                q.push(x);
              }
            }
          }
        }
      }
      if (nc[cb] < k && !removed[cb]) {
        removed[cb] = true;
        cans--;
        queue<int> q;
        q.push(cb);
        while (!q.empty()) {
          int cur = q.front();
          q.pop();
          for (int j = 0; j < adj[cur].size(); j++) {
            int x = adj[cur][j];
            if (remedges.find(make_pair(cur, x)) != remedges.end()) {
              continue;
            }
            if (!removed[x]) {
              nc[x]--;
              if (nc[x] < k) {
                removed[x] = true;
                cans--;
                q.push(x);
              }
            }
          }
        }
      }
    }
    if (cm > 0) {
      ans[cm - 1] = cans;
    }
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
}

```