#include <bits/stdc++.h>
using namespace std;
struct edge {
  int to, net;
};
int head[2000000];
int e_n;
edge edges[2000000];
int deg[2000000];
bool sd[2000000];
int v1[2000000], v2[2000000];
set<pair<int, int> > help;
set<pair<int, int> > tt;
void Init() {
  help.clear();
  e_n = 0;
  memset(sd, 0, sizeof(sd));
  memset(deg, 0, sizeof(deg));
  memset(head, -1, sizeof(head));
}
void add(int f, int t) {
  int tmp = head[f];
  edges[e_n++] = edge{t, tmp};
  head[f] = e_n - 1;
}
int main(void) {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, m, k;
  Init();
  cin >> n >> m >> k;
  int i, j, x, y;
  for (i = 1; i <= m; i++) {
    cin >> x >> y;
    v1[i] = x, v2[i] = y;
    add(x, y);
    add(y, x);
    ++deg[x], ++deg[y];
  }
  for (i = 1; i <= n; i++) {
    help.insert(make_pair(deg[i], i));
  }
  list<int> res;
  for (i = m; i >= 1; i--) {
    while (!help.empty() && help.begin()->first < k) {
      int cur = help.begin()->second;
      sd[cur] = 1;
      help.erase(help.begin());
      for (j = head[cur]; j != -1; j = edges[j].net) {
        int t = edges[j].to;
        if (!tt.count(make_pair(cur, t))) {
          help.erase(help.find(make_pair(deg[t], t)));
          deg[t]--;
          help.insert(make_pair(deg[t], t));
          tt.insert(make_pair(cur, t));
          tt.insert(make_pair(t, cur));
        }
      }
    }
    res.push_front(help.size());
    if (sd[v1[i]] == 0 && sd[v2[i]] == 0) {
      help.erase(help.find(make_pair(deg[v1[i]], v1[i])));
      help.insert(make_pair(--deg[v1[i]], v1[i]));
      help.erase(help.find(make_pair(deg[v2[i]], v2[i])));
      help.insert(make_pair(--deg[v2[i]], v2[i]));
    }
    tt.insert(make_pair(v1[i], v2[i]));
    tt.insert(make_pair(v2[i], v1[i]));
  }
  for (auto a : res) cout << a << "\n";
  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;
struct edge {
  int to, net;
};
int head[2000000];
int e_n;
edge edges[2000000];
int deg[2000000];
bool sd[2000000];
int v1[2000000], v2[2000000];
set<pair<int, int> > help;
set<pair<int, int> > tt;
void Init() {
  help.clear();
  e_n = 0;
  memset(sd, 0, sizeof(sd));
  memset(deg, 0, sizeof(deg));
  memset(head, -1, sizeof(head));
}
void add(int f, int t) {
  int tmp = head[f];
  edges[e_n++] = edge{t, tmp};
  head[f] = e_n - 1;
}
int main(void) {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, m, k;
  Init();
  cin >> n >> m >> k;
  int i, j, x, y;
  for (i = 1; i <= m; i++) {
    cin >> x >> y;
    v1[i] = x, v2[i] = y;
    add(x, y);
    add(y, x);
    ++deg[x], ++deg[y];
  }
  for (i = 1; i <= n; i++) {
    help.insert(make_pair(deg[i], i));
  }
  list<int> res;
  for (i = m; i >= 1; i--) {
    while (!help.empty() && help.begin()->first < k) {
      int cur = help.begin()->second;
      sd[cur] = 1;
      help.erase(help.begin());
      for (j = head[cur]; j != -1; j = edges[j].net) {
        int t = edges[j].to;
        if (!tt.count(make_pair(cur, t))) {
          help.erase(help.find(make_pair(deg[t], t)));
          deg[t]--;
          help.insert(make_pair(deg[t], t));
          tt.insert(make_pair(cur, t));
          tt.insert(make_pair(t, cur));
        }
      }
    }
    res.push_front(help.size());
    if (sd[v1[i]] == 0 && sd[v2[i]] == 0) {
      help.erase(help.find(make_pair(deg[v1[i]], v1[i])));
      help.insert(make_pair(--deg[v1[i]], v1[i]));
      help.erase(help.find(make_pair(deg[v2[i]], v2[i])));
      help.insert(make_pair(--deg[v2[i]], v2[i]));
    }
    tt.insert(make_pair(v1[i], v2[i]));
    tt.insert(make_pair(v2[i], v1[i]));
  }
  for (auto a : res) cout << a << "\n";
  return 0;
}

```