#include <bits/stdc++.h>
const int N = 200005;
using namespace std;
int n, m, k, deg[N], ans[N];
set<pair<int, int> > st;
vector<pair<int, int> > g[N], h;
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    h.push_back({u, v});
    g[u].push_back({v, i});
    g[v].push_back({u, i});
    deg[u]++;
    deg[v]++;
  }
  for (int i = 1; i <= n; i++) st.insert({deg[i], i});
  while (st.size() && (*(st.begin())).first < k) {
    pair<int, int> e = *st.begin();
    st.erase(e);
    int u = e.second;
    for (auto E : g[u]) {
      int el = E.first;
      if (st.count({deg[el], el})) {
        st.erase({deg[el], el});
        deg[el]--;
        st.insert({deg[el], el});
      }
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = st.size();
    int u = h[i].first, v = h[i].second;
    if (st.count({deg[u], u}) && st.count({deg[v], v})) {
      st.erase({deg[u], u});
      deg[u]--;
      st.insert({deg[u], u});
      st.erase({deg[v], v});
      deg[v]--;
      st.insert({deg[v], v});
      while (st.size() && (*(st.begin())).first < k) {
        pair<int, int> e = *st.begin();
        st.erase(e);
        int u = e.second;
        for (auto E : g[u]) {
          int el = E.first;
          if (E.second >= i) continue;
          if (st.count({deg[el], el})) {
            st.erase({deg[el], el});
            deg[el]--;
            st.insert({deg[el], el});
          }
        }
      }
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
  return 0;
}

### Prompt

Generate 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>
const int N = 200005;
using namespace std;
int n, m, k, deg[N], ans[N];
set<pair<int, int> > st;
vector<pair<int, int> > g[N], h;
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    h.push_back({u, v});
    g[u].push_back({v, i});
    g[v].push_back({u, i});
    deg[u]++;
    deg[v]++;
  }
  for (int i = 1; i <= n; i++) st.insert({deg[i], i});
  while (st.size() && (*(st.begin())).first < k) {
    pair<int, int> e = *st.begin();
    st.erase(e);
    int u = e.second;
    for (auto E : g[u]) {
      int el = E.first;
      if (st.count({deg[el], el})) {
        st.erase({deg[el], el});
        deg[el]--;
        st.insert({deg[el], el});
      }
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = st.size();
    int u = h[i].first, v = h[i].second;
    if (st.count({deg[u], u}) && st.count({deg[v], v})) {
      st.erase({deg[u], u});
      deg[u]--;
      st.insert({deg[u], u});
      st.erase({deg[v], v});
      deg[v]--;
      st.insert({deg[v], v});
      while (st.size() && (*(st.begin())).first < k) {
        pair<int, int> e = *st.begin();
        st.erase(e);
        int u = e.second;
        for (auto E : g[u]) {
          int el = E.first;
          if (E.second >= i) continue;
          if (st.count({deg[el], el})) {
            st.erase({deg[el], el});
            deg[el]--;
            st.insert({deg[el], el});
          }
        }
      }
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
  return 0;
}

```