#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const long long MOD = 1e9 + 7;
pair<int, int> es[200000];
vector<pair<int, int> > g[200000];
int degree[200000];
set<pair<int, int> > st;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (long long i = 0; i < m; i++) {
    scanf("%d%d", &es[i].first, &es[i].second);
    es[i].first--, es[i].second--;
    g[es[i].first].push_back({es[i].second, i});
    g[es[i].second].push_back({es[i].first, i});
    degree[es[i].first]++, degree[es[i].second]++;
  }
  reverse(es, es + m);
  for (long long i = 0; i < n; i++) st.insert({degree[i], i});
  while (!st.empty() && st.begin()->first < k) {
    auto v = st.begin();
    for (pair<int, int> &i : g[v->second]) {
      int to = i.first;
      if (st.find({degree[to], to}) != st.end()) {
        st.erase({degree[to], to});
        degree[to]--;
        st.insert({degree[to], to});
      }
    }
    st.erase(v);
  }
  vector<int> out;
  for (long long j = 0; j < m; j++) {
    out.push_back(st.size());
    int u = es[j].first, v = es[j].second;
    if (st.find({degree[u], u}) != st.end() &&
        st.find({degree[v], v}) != st.end()) {
      st.erase({degree[u], u});
      degree[u]--;
      st.insert({degree[u], u});
      st.erase({degree[v], v});
      degree[v]--;
      st.insert({degree[v], v});
      while (!st.empty() && st.begin()->first < k) {
        auto v = st.begin();
        for (pair<int, int> &i : g[v->second]) {
          if (i.second >= m - j - 1) continue;
          int to = i.first;
          if (st.find({degree[to], to}) != st.end()) {
            st.erase({degree[to], to});
            degree[to]--;
            st.insert({degree[to], to});
          }
        }
        st.erase(v);
      }
    }
  }
  reverse(begin(out), end(out));
  for (int i : out) printf("%d\n", i);
}

### 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 double EPS = 1e-9;
const long long MOD = 1e9 + 7;
pair<int, int> es[200000];
vector<pair<int, int> > g[200000];
int degree[200000];
set<pair<int, int> > st;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (long long i = 0; i < m; i++) {
    scanf("%d%d", &es[i].first, &es[i].second);
    es[i].first--, es[i].second--;
    g[es[i].first].push_back({es[i].second, i});
    g[es[i].second].push_back({es[i].first, i});
    degree[es[i].first]++, degree[es[i].second]++;
  }
  reverse(es, es + m);
  for (long long i = 0; i < n; i++) st.insert({degree[i], i});
  while (!st.empty() && st.begin()->first < k) {
    auto v = st.begin();
    for (pair<int, int> &i : g[v->second]) {
      int to = i.first;
      if (st.find({degree[to], to}) != st.end()) {
        st.erase({degree[to], to});
        degree[to]--;
        st.insert({degree[to], to});
      }
    }
    st.erase(v);
  }
  vector<int> out;
  for (long long j = 0; j < m; j++) {
    out.push_back(st.size());
    int u = es[j].first, v = es[j].second;
    if (st.find({degree[u], u}) != st.end() &&
        st.find({degree[v], v}) != st.end()) {
      st.erase({degree[u], u});
      degree[u]--;
      st.insert({degree[u], u});
      st.erase({degree[v], v});
      degree[v]--;
      st.insert({degree[v], v});
      while (!st.empty() && st.begin()->first < k) {
        auto v = st.begin();
        for (pair<int, int> &i : g[v->second]) {
          if (i.second >= m - j - 1) continue;
          int to = i.first;
          if (st.find({degree[to], to}) != st.end()) {
            st.erase({degree[to], to});
            degree[to]--;
            st.insert({degree[to], to});
          }
        }
        st.erase(v);
      }
    }
  }
  reverse(begin(out), end(out));
  for (int i : out) printf("%d\n", i);
}

```