#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int deg[N], del[N], u[N], v[N], aa[N], edg[N];
vector<pair<int, int> > adj[N];
int remove(queue<int> &q, int k) {
  int ans = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (!del[x]) del[x] = 1, ans--;
    for (auto z : adj[x]) {
      int idx = z.second, y = z.first;
      if (edg[idx]) continue;
      edg[idx] = 1;
      deg[x]--;
      deg[y]--;
      if (deg[y] < k && !del[y]) {
        q.push(y);
      }
    }
  }
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> u[i] >> v[i];
    adj[u[i]].push_back({v[i], i});
    adj[v[i]].push_back({u[i], i});
    deg[u[i]]++;
    deg[v[i]]++;
  }
  int ans = n;
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) q.push(i);
  ans += remove(q, k);
  for (int i = m; i >= 1; i--) {
    aa[i] = ans;
    if (edg[i]) continue;
    deg[u[i]]--;
    deg[v[i]]--;
    edg[i] = 1;
    if (!del[u[i]] && deg[u[i]] < k) q.push(u[i]);
    if (!del[v[i]] && deg[v[i]] < k) q.push(v[i]);
    ans += remove(q, k);
  }
  for (int i = 1; i <= m; i++) cout << aa[i] << "\n";
}

### 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;
const int N = 2e5 + 7;
int deg[N], del[N], u[N], v[N], aa[N], edg[N];
vector<pair<int, int> > adj[N];
int remove(queue<int> &q, int k) {
  int ans = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (!del[x]) del[x] = 1, ans--;
    for (auto z : adj[x]) {
      int idx = z.second, y = z.first;
      if (edg[idx]) continue;
      edg[idx] = 1;
      deg[x]--;
      deg[y]--;
      if (deg[y] < k && !del[y]) {
        q.push(y);
      }
    }
  }
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> u[i] >> v[i];
    adj[u[i]].push_back({v[i], i});
    adj[v[i]].push_back({u[i], i});
    deg[u[i]]++;
    deg[v[i]]++;
  }
  int ans = n;
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) q.push(i);
  ans += remove(q, k);
  for (int i = m; i >= 1; i--) {
    aa[i] = ans;
    if (edg[i]) continue;
    deg[u[i]]--;
    deg[v[i]]--;
    edg[i] = 1;
    if (!del[u[i]] && deg[u[i]] < k) q.push(u[i]);
    if (!del[v[i]] && deg[v[i]] < k) q.push(v[i]);
    ans += remove(q, k);
  }
  for (int i = 1; i <= m; i++) cout << aa[i] << "\n";
}

```