#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5;
set<int> g[MAXN];
int n, m, k;
int result = 0;
bool removed[MAXN];
int c[MAXN];
int edges[MAXN][2];
queue<int> q;
void remove_item(int m) {
  result--;
  q.push(m);
  removed[m] = true;
}
void clear_queue() {
  while (!q.empty()) {
    auto to = q.front();
    q.pop();
    for (auto u : g[to]) {
      c[u]--;
      if (c[u] < k && !removed[u]) remove_item(u);
    }
  }
}
void remove_edge(int a, int b) {
  g[a].erase(b);
  g[b].erase(a);
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    edges[i][0] = a;
    edges[i][1] = b;
    g[a].insert(b);
    g[b].insert(a);
    c[a]++;
    c[b]++;
  }
  for (int i = 0; i < n; i++) {
    if (c[i] < k) remove_item(i);
  }
  clear_queue();
  vector<int> result_reversed;
  result_reversed.push_back(n + result);
  for (int i = m - 1; i > 0; i--) {
    int a = edges[i][0];
    int b = edges[i][1];
    if (removed[a] || removed[b]) {
      result_reversed.push_back(n + result);
      continue;
    }
    c[a]--;
    c[b]--;
    g[a].erase(b);
    g[b].erase(a);
    if (c[a] < k) remove_item(a);
    if (c[b] < k && !removed[b]) remove_item(b);
    clear_queue();
    result_reversed.push_back(n + result);
  }
  for (int i = m - 1; i >= 0; i--) {
    printf("%d\n", result_reversed[i]);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
set<int> g[MAXN];
int n, m, k;
int result = 0;
bool removed[MAXN];
int c[MAXN];
int edges[MAXN][2];
queue<int> q;
void remove_item(int m) {
  result--;
  q.push(m);
  removed[m] = true;
}
void clear_queue() {
  while (!q.empty()) {
    auto to = q.front();
    q.pop();
    for (auto u : g[to]) {
      c[u]--;
      if (c[u] < k && !removed[u]) remove_item(u);
    }
  }
}
void remove_edge(int a, int b) {
  g[a].erase(b);
  g[b].erase(a);
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    edges[i][0] = a;
    edges[i][1] = b;
    g[a].insert(b);
    g[b].insert(a);
    c[a]++;
    c[b]++;
  }
  for (int i = 0; i < n; i++) {
    if (c[i] < k) remove_item(i);
  }
  clear_queue();
  vector<int> result_reversed;
  result_reversed.push_back(n + result);
  for (int i = m - 1; i > 0; i--) {
    int a = edges[i][0];
    int b = edges[i][1];
    if (removed[a] || removed[b]) {
      result_reversed.push_back(n + result);
      continue;
    }
    c[a]--;
    c[b]--;
    g[a].erase(b);
    g[b].erase(a);
    if (c[a] < k) remove_item(a);
    if (c[b] < k && !removed[b]) remove_item(b);
    clear_queue();
    result_reversed.push_back(n + result);
  }
  for (int i = m - 1; i >= 0; i--) {
    printf("%d\n", result_reversed[i]);
  }
  return 0;
}

```