#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
template <typename T>
void DBG(const char* name, T&& H) {
  cerr << name << " = " << H << ')' << '\n';
}
template <typename T, typename... Args>
void DBG(const char* names, T&& H, Args&&... args) {
  const char* NEXT = strchr(names + 1, ',');
  cerr.write(names, NEXT - names) << " = " << H << " |";
  DBG(NEXT + 1, args...);
}
using ll = long long;
using ld = long double;
const ll mod1 = 1e9 + 7;
const ld PI = acos(-1.0);
const ll maxN = 1e6 + 1;
const ll INF = 1e18;
void Solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> deg(n + 1, 0);
  vector<pair<int, int>> friendship(m);
  vector<set<int>> graph(n + 1);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    deg[a]++, deg[b]++;
    graph[a].insert(b);
    graph[b].insert(a);
    friendship[i] = {a, b};
  }
  set<pair<int, int>> q;
  for (int i = 1; i <= n; i++) {
    q.insert({deg[i], i});
  }
  vector<int> ans(m, 0);
  for (int i = m - 1; i >= 0; i--) {
    while (q.size() && ((*q.begin()).first < k)) {
      int ver = (*q.begin()).second;
      for (auto j : graph[ver]) {
        graph[j].erase(ver);
        q.erase({deg[j], j});
        q.insert({deg[j] - 1, j});
        deg[j]--;
      }
      q.erase({deg[ver], ver});
    }
    ans[i] = q.size();
    int a = friendship[i].first;
    int b = friendship[i].second;
    if (q.count({deg[a], a}) && q.count({deg[b], b})) {
      q.erase({deg[a], a});
      q.erase({deg[b], b});
      graph[a].erase(b);
      graph[b].erase(a);
      deg[a]--;
      deg[b]--;
      q.insert({deg[a], a});
      q.insert({deg[b], b});
    }
  }
  for (auto i : ans) cout << i << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  while (tt--) {
    Solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
template <typename T>
void DBG(const char* name, T&& H) {
  cerr << name << " = " << H << ')' << '\n';
}
template <typename T, typename... Args>
void DBG(const char* names, T&& H, Args&&... args) {
  const char* NEXT = strchr(names + 1, ',');
  cerr.write(names, NEXT - names) << " = " << H << " |";
  DBG(NEXT + 1, args...);
}
using ll = long long;
using ld = long double;
const ll mod1 = 1e9 + 7;
const ld PI = acos(-1.0);
const ll maxN = 1e6 + 1;
const ll INF = 1e18;
void Solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> deg(n + 1, 0);
  vector<pair<int, int>> friendship(m);
  vector<set<int>> graph(n + 1);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    deg[a]++, deg[b]++;
    graph[a].insert(b);
    graph[b].insert(a);
    friendship[i] = {a, b};
  }
  set<pair<int, int>> q;
  for (int i = 1; i <= n; i++) {
    q.insert({deg[i], i});
  }
  vector<int> ans(m, 0);
  for (int i = m - 1; i >= 0; i--) {
    while (q.size() && ((*q.begin()).first < k)) {
      int ver = (*q.begin()).second;
      for (auto j : graph[ver]) {
        graph[j].erase(ver);
        q.erase({deg[j], j});
        q.insert({deg[j] - 1, j});
        deg[j]--;
      }
      q.erase({deg[ver], ver});
    }
    ans[i] = q.size();
    int a = friendship[i].first;
    int b = friendship[i].second;
    if (q.count({deg[a], a}) && q.count({deg[b], b})) {
      q.erase({deg[a], a});
      q.erase({deg[b], b});
      graph[a].erase(b);
      graph[b].erase(a);
      deg[a]--;
      deg[b]--;
      q.insert({deg[a], a});
      q.insert({deg[b], b});
    }
  }
  for (auto i : ans) cout << i << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  while (tt--) {
    Solve();
  }
  return 0;
}

```