#include <bits/stdc++.h>
using namespace std;
int n, m, k, x[200009], y[200009];
set<int> alive;
vector<int> kaaj;
set<int> graph[200009];
int degree[200009], ans[200009];
void doit(int z) {
  if (alive.find(z) == alive.end()) return;
  alive.erase(z);
  if (graph[z].empty()) return;
  auto it = graph[z].begin();
  while (it != graph[z].end()) {
    int v = *it;
    degree[v]--;
    graph[v].erase(z);
    if (degree[v] < k && alive.find(v) != alive.end()) {
      kaaj.push_back(v);
    }
    it++;
  }
  graph[z].clear();
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &x[i], &y[i]);
    graph[x[i]].insert(y[i]);
    graph[y[i]].insert(x[i]);
    degree[x[i]]++;
    degree[y[i]]++;
  }
  for (int i = 1; i <= n; i++) alive.insert(i);
  for (int i = 1; i <= n; i++) {
    if (degree[i] < k) {
      kaaj.push_back(i);
    }
  }
  while (!kaaj.empty()) {
    int z = kaaj.back();
    kaaj.pop_back();
    doit(z);
  }
  for (int i = m; i >= 1; i--) {
    ans[i] = alive.size();
    if (alive.find(x[i]) == alive.end() || alive.find(y[i]) == alive.end())
      continue;
    degree[x[i]]--;
    degree[y[i]]--;
    graph[x[i]].erase(y[i]);
    graph[y[i]].erase(x[i]);
    if (degree[x[i]] < k && alive.find(x[i]) != alive.end()) {
      kaaj.push_back(x[i]);
    }
    while (!kaaj.empty()) {
      int z = kaaj.back();
      kaaj.pop_back();
      doit(z);
    }
    if (degree[y[i]] < k && alive.find(y[i]) != alive.end()) {
      kaaj.push_back(y[i]);
    }
    while (!kaaj.empty()) {
      int z = kaaj.back();
      kaaj.pop_back();
      doit(z);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int n, m, k, x[200009], y[200009];
set<int> alive;
vector<int> kaaj;
set<int> graph[200009];
int degree[200009], ans[200009];
void doit(int z) {
  if (alive.find(z) == alive.end()) return;
  alive.erase(z);
  if (graph[z].empty()) return;
  auto it = graph[z].begin();
  while (it != graph[z].end()) {
    int v = *it;
    degree[v]--;
    graph[v].erase(z);
    if (degree[v] < k && alive.find(v) != alive.end()) {
      kaaj.push_back(v);
    }
    it++;
  }
  graph[z].clear();
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &x[i], &y[i]);
    graph[x[i]].insert(y[i]);
    graph[y[i]].insert(x[i]);
    degree[x[i]]++;
    degree[y[i]]++;
  }
  for (int i = 1; i <= n; i++) alive.insert(i);
  for (int i = 1; i <= n; i++) {
    if (degree[i] < k) {
      kaaj.push_back(i);
    }
  }
  while (!kaaj.empty()) {
    int z = kaaj.back();
    kaaj.pop_back();
    doit(z);
  }
  for (int i = m; i >= 1; i--) {
    ans[i] = alive.size();
    if (alive.find(x[i]) == alive.end() || alive.find(y[i]) == alive.end())
      continue;
    degree[x[i]]--;
    degree[y[i]]--;
    graph[x[i]].erase(y[i]);
    graph[y[i]].erase(x[i]);
    if (degree[x[i]] < k && alive.find(x[i]) != alive.end()) {
      kaaj.push_back(x[i]);
    }
    while (!kaaj.empty()) {
      int z = kaaj.back();
      kaaj.pop_back();
      doit(z);
    }
    if (degree[y[i]] < k && alive.find(y[i]) != alive.end()) {
      kaaj.push_back(y[i]);
    }
    while (!kaaj.empty()) {
      int z = kaaj.back();
      kaaj.pop_back();
      doit(z);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```