#include <bits/stdc++.h>
using namespace std;
const int max_N = 2e5 + 10;
int x[max_N], y[max_N];
set<int> E[max_N];
void remove(int u);
int N, M, K;
int ans = 0;
int active[max_N];
int main() {
  cin >> N >> M >> K;
  for (int i = 1; i <= M; ++i) {
    cin >> x[i] >> y[i];
    E[x[i]].insert(y[i]);
    E[y[i]].insert(x[i]);
  }
  for (int i = 1; i <= N; ++i) {
    active[i] = 1;
  }
  ans = N;
  for (int i = 1; i <= N; ++i) {
    remove(i);
  }
  vector<int> ans;
  for (int i = M; i >= 1; --i) {
    ans.push_back(::ans);
    int u = x[i], v = y[i];
    E[u].erase(v);
    E[v].erase(u);
    remove(u);
    remove(v);
  }
  reverse(ans.begin(), ans.end());
  for (auto &x : ans) {
    cout << x << '\n';
  }
}
void remove(int u) {
  if (E[u].size() >= K || !active[u]) {
    return;
  }
  active[u] = 0;
  ans--;
  for (auto &v : E[u]) {
    E[v].erase(u);
    remove(v);
  }
}

### 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>
using namespace std;
const int max_N = 2e5 + 10;
int x[max_N], y[max_N];
set<int> E[max_N];
void remove(int u);
int N, M, K;
int ans = 0;
int active[max_N];
int main() {
  cin >> N >> M >> K;
  for (int i = 1; i <= M; ++i) {
    cin >> x[i] >> y[i];
    E[x[i]].insert(y[i]);
    E[y[i]].insert(x[i]);
  }
  for (int i = 1; i <= N; ++i) {
    active[i] = 1;
  }
  ans = N;
  for (int i = 1; i <= N; ++i) {
    remove(i);
  }
  vector<int> ans;
  for (int i = M; i >= 1; --i) {
    ans.push_back(::ans);
    int u = x[i], v = y[i];
    E[u].erase(v);
    E[v].erase(u);
    remove(u);
    remove(v);
  }
  reverse(ans.begin(), ans.end());
  for (auto &x : ans) {
    cout << x << '\n';
  }
}
void remove(int u) {
  if (E[u].size() >= K || !active[u]) {
    return;
  }
  active[u] = 0;
  ans--;
  for (auto &v : E[u]) {
    E[v].erase(u);
    remove(v);
  }
}

```