#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int N, M, K;
int x[MAXN], y[MAXN];
unordered_set<int> adj[MAXN];
int ans[MAXN];
struct cmp {
  bool operator()(const int &lhs, const int &rhs) const {
    if (adj[lhs].size() != adj[rhs].size())
      return adj[lhs].size() < adj[rhs].size();
    return lhs < rhs;
  }
};
set<int, cmp> in;
void load() {
  scanf("%d%d%d", &N, &M, &K);
  for (int i = 0; i < M; i++) scanf("%d%d", x + i, y + i);
}
void reduce() {
  if (in.empty()) return;
  int x = *in.begin();
  if (adj[x].size() >= K) return;
  in.erase(x);
  for (auto it : adj[x])
    if (in.count(it)) {
      in.erase(it);
      adj[it].erase(x);
      in.insert(it);
    }
  reduce();
}
void solve() {
  for (int i = 0; i < M; i++) {
    adj[x[i]].insert(y[i]);
    adj[y[i]].insert(x[i]);
  }
  for (int i = 1; i <= N; i++) in.insert(i);
  reduce();
  for (int i = M - 1; i >= 0; i--) {
    ans[i] = in.size();
    if (in.count(x[i]) && in.count(y[i])) {
      in.erase(x[i]);
      in.erase(y[i]);
      adj[x[i]].erase(y[i]);
      adj[y[i]].erase(x[i]);
      in.insert(x[i]);
      in.insert(y[i]);
      reduce();
    }
  }
  for (int i = 0; i < M; i++) printf("%d\n", ans[i]);
}
int main() {
  load();
  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>
using namespace std;
const int MAXN = 2e5 + 5;
int N, M, K;
int x[MAXN], y[MAXN];
unordered_set<int> adj[MAXN];
int ans[MAXN];
struct cmp {
  bool operator()(const int &lhs, const int &rhs) const {
    if (adj[lhs].size() != adj[rhs].size())
      return adj[lhs].size() < adj[rhs].size();
    return lhs < rhs;
  }
};
set<int, cmp> in;
void load() {
  scanf("%d%d%d", &N, &M, &K);
  for (int i = 0; i < M; i++) scanf("%d%d", x + i, y + i);
}
void reduce() {
  if (in.empty()) return;
  int x = *in.begin();
  if (adj[x].size() >= K) return;
  in.erase(x);
  for (auto it : adj[x])
    if (in.count(it)) {
      in.erase(it);
      adj[it].erase(x);
      in.insert(it);
    }
  reduce();
}
void solve() {
  for (int i = 0; i < M; i++) {
    adj[x[i]].insert(y[i]);
    adj[y[i]].insert(x[i]);
  }
  for (int i = 1; i <= N; i++) in.insert(i);
  reduce();
  for (int i = M - 1; i >= 0; i--) {
    ans[i] = in.size();
    if (in.count(x[i]) && in.count(y[i])) {
      in.erase(x[i]);
      in.erase(y[i]);
      adj[x[i]].erase(y[i]);
      adj[y[i]].erase(x[i]);
      in.insert(x[i]);
      in.insert(y[i]);
      reduce();
    }
  }
  for (int i = 0; i < M; i++) printf("%d\n", ans[i]);
}
int main() {
  load();
  solve();
  return 0;
}

```