#include <bits/stdc++.h>
using namespace std;
const int nmax = 2e5 + 5;
const int mod = 1e9 + 7;
vector<int> g[nmax];
int destroyed;
pair<int, int> ej[nmax];
int deg[nmax], ans[nmax];
bool vis[nmax];
map<pair<int, int>, int> mp;
void dhongsho(int u, int k, int p) {
  if (vis[u]) return;
  vis[u] = true;
  destroyed++;
  for (auto v : g[u]) {
    if (v == p or vis[v]) continue;
    if (mp[{u, v}] == 0) continue;
    deg[v]--;
    if (deg[v] < k) dhongsho(v, k, u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k, u, v;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    g[u].push_back(v), g[v].push_back(u);
    ej[i] = {u, v};
    mp[{u, v}] = mp[{v, u}] = 1;
    deg[u]++, deg[v]++;
  }
  for (int u = 1; u <= n; u++) {
    if (deg[u] < k) dhongsho(u, k, -1);
  }
  for (int i = m; i >= 1; i--) {
    ans[i] = n - destroyed;
    u = ej[i].first, v = ej[i].second;
    if (!vis[u] and !vis[v]) {
      deg[u]--, deg[v]--;
      mp[{u, v}] = mp[{v, u}] = 0;
      if (deg[u] < k) dhongsho(u, k, v);
      if (deg[v] < k) dhongsho(v, k, u);
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 nmax = 2e5 + 5;
const int mod = 1e9 + 7;
vector<int> g[nmax];
int destroyed;
pair<int, int> ej[nmax];
int deg[nmax], ans[nmax];
bool vis[nmax];
map<pair<int, int>, int> mp;
void dhongsho(int u, int k, int p) {
  if (vis[u]) return;
  vis[u] = true;
  destroyed++;
  for (auto v : g[u]) {
    if (v == p or vis[v]) continue;
    if (mp[{u, v}] == 0) continue;
    deg[v]--;
    if (deg[v] < k) dhongsho(v, k, u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k, u, v;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    g[u].push_back(v), g[v].push_back(u);
    ej[i] = {u, v};
    mp[{u, v}] = mp[{v, u}] = 1;
    deg[u]++, deg[v]++;
  }
  for (int u = 1; u <= n; u++) {
    if (deg[u] < k) dhongsho(u, k, -1);
  }
  for (int i = m; i >= 1; i--) {
    ans[i] = n - destroyed;
    u = ej[i].first, v = ej[i].second;
    if (!vis[u] and !vis[v]) {
      deg[u]--, deg[v]--;
      mp[{u, v}] = mp[{v, u}] = 0;
      if (deg[u] < k) dhongsho(u, k, v);
      if (deg[v] < k) dhongsho(v, k, u);
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}

```