#include <bits/stdc++.h>
using namespace std;
int n, m, k;
int a[200010], b[200010];
vector<pair<int, int> > G[200010];
int deg[200010];
int ans[200010], cnt;
queue<int> q;
inline void spfa(int x) {
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i].first;
      int s = G[u][i].second;
      if (s >= x) continue;
      if (deg[v] >= k) {
        deg[v]--;
        if (deg[v] < k) {
          q.push(v);
          cnt--;
        }
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", a + i, b + i);
    deg[a[i]]++;
    deg[b[i]]++;
    G[a[i]].push_back(pair<int, int>(b[i], i));
    G[b[i]].push_back(pair<int, int>(a[i], i));
  }
  cnt = n;
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) q.push(i), cnt--;
  spfa(m + 1);
  ans[m + 1] = cnt;
  for (int i = m; i > 1; i--) {
    if (deg[a[i]] >= k && deg[b[i]] >= k) {
      deg[a[i]]--;
      deg[b[i]]--;
      if (deg[a[i]] < k) q.push(a[i]), cnt--;
      if (deg[b[i]] < k) q.push(b[i]), cnt--;
    }
    spfa(i);
    ans[i] = cnt;
  }
  for (int i = 2; i <= m + 1; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your task is to create 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 a[200010], b[200010];
vector<pair<int, int> > G[200010];
int deg[200010];
int ans[200010], cnt;
queue<int> q;
inline void spfa(int x) {
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i].first;
      int s = G[u][i].second;
      if (s >= x) continue;
      if (deg[v] >= k) {
        deg[v]--;
        if (deg[v] < k) {
          q.push(v);
          cnt--;
        }
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", a + i, b + i);
    deg[a[i]]++;
    deg[b[i]]++;
    G[a[i]].push_back(pair<int, int>(b[i], i));
    G[b[i]].push_back(pair<int, int>(a[i], i));
  }
  cnt = n;
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) q.push(i), cnt--;
  spfa(m + 1);
  ans[m + 1] = cnt;
  for (int i = m; i > 1; i--) {
    if (deg[a[i]] >= k && deg[b[i]] >= k) {
      deg[a[i]]--;
      deg[b[i]]--;
      if (deg[a[i]] < k) q.push(a[i]), cnt--;
      if (deg[b[i]] < k) q.push(b[i]), cnt--;
    }
    spfa(i);
    ans[i] = cnt;
  }
  for (int i = 2; i <= m + 1; i++) printf("%d\n", ans[i]);
  return 0;
}

```