#include <bits/stdc++.h>
using namespace std;
const int max_n = 200005;
struct edge {
  int u, v, next;
} G[max_n * 2];
int head[max_n];
int total;
bool visit[max_n * 2];
int in[max_n];
set<pair<int, int> > s;
int n, m, k;
int u, v;
int ans[max_n];
void add_edge(int u, int v) {
  G[total].u = u;
  G[total].v = v;
  G[total].next = head[u];
  head[u] = total++;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  total = 0;
  memset(head, -1, sizeof(head));
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
    in[u]++;
    in[v]++;
  }
  for (int i = 1; i <= n; i++) s.insert(pair<int, int>(in[i], i));
  pair<int, int> p;
  for (int i = 2 * m - 1; i >= 0; i -= 2) {
    while (!s.empty()) {
      p = *s.begin();
      if (p.first < k) {
        s.erase(p);
        for (int j = head[p.second]; ~j; j = G[j].next) {
          if (!visit[j]) {
            v = G[j].v;
            s.erase(pair<int, int>(in[v], v));
            in[v]--;
            s.insert(pair<int, int>(in[v], v));
            visit[j] = visit[j ^ 1] = true;
          }
        }
      } else
        break;
    }
    ans[i / 2] = s.size();
    if (!visit[i]) {
      u = G[i].u;
      v = G[i].v;
      s.erase(pair<int, int>(in[u], u));
      in[u]--;
      s.insert(pair<int, int>(in[u], u));
      s.erase(pair<int, int>(in[v], v));
      in[v]--;
      s.insert(pair<int, int>(in[v], v));
      visit[i] = visit[i ^ 1] = true;
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 200005;
struct edge {
  int u, v, next;
} G[max_n * 2];
int head[max_n];
int total;
bool visit[max_n * 2];
int in[max_n];
set<pair<int, int> > s;
int n, m, k;
int u, v;
int ans[max_n];
void add_edge(int u, int v) {
  G[total].u = u;
  G[total].v = v;
  G[total].next = head[u];
  head[u] = total++;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  total = 0;
  memset(head, -1, sizeof(head));
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
    in[u]++;
    in[v]++;
  }
  for (int i = 1; i <= n; i++) s.insert(pair<int, int>(in[i], i));
  pair<int, int> p;
  for (int i = 2 * m - 1; i >= 0; i -= 2) {
    while (!s.empty()) {
      p = *s.begin();
      if (p.first < k) {
        s.erase(p);
        for (int j = head[p.second]; ~j; j = G[j].next) {
          if (!visit[j]) {
            v = G[j].v;
            s.erase(pair<int, int>(in[v], v));
            in[v]--;
            s.insert(pair<int, int>(in[v], v));
            visit[j] = visit[j ^ 1] = true;
          }
        }
      } else
        break;
    }
    ans[i / 2] = s.size();
    if (!visit[i]) {
      u = G[i].u;
      v = G[i].v;
      s.erase(pair<int, int>(in[u], u));
      in[u]--;
      s.insert(pair<int, int>(in[u], u));
      s.erase(pair<int, int>(in[v], v));
      in[v]--;
      s.insert(pair<int, int>(in[v], v));
      visit[i] = visit[i ^ 1] = true;
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

```