#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
int N, M, K;
int A[MAXN];
int B[MAXN];
set<int> adj[MAXN];
int deg[MAXN];
set<pair<int, int> > st;
bool inGraph[MAXN];
int ans[MAXN];
void update(int x) {
  if (!inGraph[x]) return;
  set<pair<int, int> >::iterator it = st.find(make_pair(deg[x], x));
  st.erase(it);
  deg[x]--;
  st.insert(make_pair(deg[x], x));
}
void fix() {
  while (!st.empty() && st.begin()->first < K) {
    int cur = st.begin()->second;
    st.erase(st.begin());
    inGraph[cur] = false;
    for (auto nxt : adj[cur]) {
      adj[nxt].erase(cur);
      update(nxt);
    }
    adj[cur].clear();
  }
}
int main() {
  scanf("%d %d %d", &N, &M, &K);
  for (int i = 0; i < M; i++) {
    scanf("%d %d", &A[i], &B[i]);
    deg[A[i]]++;
    deg[B[i]]++;
    adj[A[i]].insert(B[i]);
    adj[B[i]].insert(A[i]);
  }
  for (int i = 1; i <= N; i++) {
    st.insert(make_pair(deg[i], i));
    inGraph[i] = true;
  }
  for (int i = M - 1; i >= 0; i--) {
    fix();
    ans[i] = st.size();
    adj[A[i]].erase(B[i]);
    adj[B[i]].erase(A[i]);
    if (inGraph[B[i]]) update(A[i]);
    if (inGraph[A[i]]) update(B[i]);
  }
  for (int i = 0; i < M; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const int MAXN = 200005;
int N, M, K;
int A[MAXN];
int B[MAXN];
set<int> adj[MAXN];
int deg[MAXN];
set<pair<int, int> > st;
bool inGraph[MAXN];
int ans[MAXN];
void update(int x) {
  if (!inGraph[x]) return;
  set<pair<int, int> >::iterator it = st.find(make_pair(deg[x], x));
  st.erase(it);
  deg[x]--;
  st.insert(make_pair(deg[x], x));
}
void fix() {
  while (!st.empty() && st.begin()->first < K) {
    int cur = st.begin()->second;
    st.erase(st.begin());
    inGraph[cur] = false;
    for (auto nxt : adj[cur]) {
      adj[nxt].erase(cur);
      update(nxt);
    }
    adj[cur].clear();
  }
}
int main() {
  scanf("%d %d %d", &N, &M, &K);
  for (int i = 0; i < M; i++) {
    scanf("%d %d", &A[i], &B[i]);
    deg[A[i]]++;
    deg[B[i]]++;
    adj[A[i]].insert(B[i]);
    adj[B[i]].insert(A[i]);
  }
  for (int i = 1; i <= N; i++) {
    st.insert(make_pair(deg[i], i));
    inGraph[i] = true;
  }
  for (int i = M - 1; i >= 0; i--) {
    fix();
    ans[i] = st.size();
    adj[A[i]].erase(B[i]);
    adj[B[i]].erase(A[i]);
    if (inGraph[B[i]]) update(A[i]);
    if (inGraph[A[i]]) update(B[i]);
  }
  for (int i = 0; i < M; i++) printf("%d\n", ans[i]);
  return 0;
}

```