#include <bits/stdc++.h>
using namespace std;
int N, M, K;
int xList[200005], yList[200005];
int deg[200005], ans[200005];
set<int> Adj[200005];
set<pair<int, int> > s;
set<pair<int, int> >::iterator p;
void work(int u) {
  s.erase(make_pair(deg[u], u));
  int v;
  set<int>::iterator k;
  for (k = Adj[u].begin(); k != Adj[u].end(); k++) {
    v = (*k);
    if (!s.count(make_pair(deg[v], v))) {
      deg[v]--;
      continue;
    }
    s.erase(make_pair(deg[v], v));
    deg[v]--;
    if (deg[v] >= K)
      s.insert(make_pair(deg[v], v));
    else
      work(v);
  }
}
int main() {
  scanf("%d%d%d", &N, &M, &K);
  int u, v;
  for (int i = 1; i <= M; i++) {
    scanf("%d%d", &u, &v);
    Adj[u].insert(v);
    Adj[v].insert(u);
    deg[u]++;
    deg[v]++;
    xList[i] = u;
    yList[i] = v;
  }
  for (int i = 1; i <= N; i++) {
    s.insert(make_pair(deg[i], i));
  }
  while (!s.empty()) {
    p = s.begin();
    if (p->first < K)
      work(p->second);
    else
      break;
  }
  ans[M] = s.size();
  for (int i = M; i >= 1; i--) {
    u = xList[i];
    v = yList[i];
    ans[i - 1] = s.size();
    if (!s.count(make_pair(deg[u], u)) || !s.count(make_pair(deg[v], v)))
      continue;
    Adj[u].erase(v);
    Adj[v].erase(u);
    if (s.count(make_pair(deg[u], u))) {
      s.erase(make_pair(deg[u], u));
      deg[u]--;
      s.insert(make_pair(deg[u], u));
    }
    if (s.count(make_pair(deg[v], v))) {
      s.erase(make_pair(deg[v], v));
      deg[v]--;
      s.insert(make_pair(deg[v], v));
    }
    while (!s.empty()) {
      p = s.begin();
      if (p->first < K)
        work(p->second);
      else
        break;
    }
    ans[i - 1] = s.size();
  }
  for (int i = 1; i <= M; i++) printf("%d\n", ans[i]);
  getchar();
  return 0;
}

### Prompt

Please formulate 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;
int N, M, K;
int xList[200005], yList[200005];
int deg[200005], ans[200005];
set<int> Adj[200005];
set<pair<int, int> > s;
set<pair<int, int> >::iterator p;
void work(int u) {
  s.erase(make_pair(deg[u], u));
  int v;
  set<int>::iterator k;
  for (k = Adj[u].begin(); k != Adj[u].end(); k++) {
    v = (*k);
    if (!s.count(make_pair(deg[v], v))) {
      deg[v]--;
      continue;
    }
    s.erase(make_pair(deg[v], v));
    deg[v]--;
    if (deg[v] >= K)
      s.insert(make_pair(deg[v], v));
    else
      work(v);
  }
}
int main() {
  scanf("%d%d%d", &N, &M, &K);
  int u, v;
  for (int i = 1; i <= M; i++) {
    scanf("%d%d", &u, &v);
    Adj[u].insert(v);
    Adj[v].insert(u);
    deg[u]++;
    deg[v]++;
    xList[i] = u;
    yList[i] = v;
  }
  for (int i = 1; i <= N; i++) {
    s.insert(make_pair(deg[i], i));
  }
  while (!s.empty()) {
    p = s.begin();
    if (p->first < K)
      work(p->second);
    else
      break;
  }
  ans[M] = s.size();
  for (int i = M; i >= 1; i--) {
    u = xList[i];
    v = yList[i];
    ans[i - 1] = s.size();
    if (!s.count(make_pair(deg[u], u)) || !s.count(make_pair(deg[v], v)))
      continue;
    Adj[u].erase(v);
    Adj[v].erase(u);
    if (s.count(make_pair(deg[u], u))) {
      s.erase(make_pair(deg[u], u));
      deg[u]--;
      s.insert(make_pair(deg[u], u));
    }
    if (s.count(make_pair(deg[v], v))) {
      s.erase(make_pair(deg[v], v));
      deg[v]--;
      s.insert(make_pair(deg[v], v));
    }
    while (!s.empty()) {
      p = s.begin();
      if (p->first < K)
        work(p->second);
      else
        break;
    }
    ans[i - 1] = s.size();
  }
  for (int i = 1; i <= M; i++) printf("%d\n", ans[i]);
  getchar();
  return 0;
}

```