#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 3;
int n, m, k, deg[N], x[N], y[N], ans[N], cnt;
bool del[N];
set<int> G[N];
set<int>::iterator it;
void Del(int rt) {
  if (deg[rt] >= k || del[rt]) return;
  queue<int> q;
  q.push(rt);
  del[rt] = 1;
  cnt--;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (it = G[u].begin(); it != G[u].end(); it++) {
      int v = *it;
      deg[v]--;
      if (deg[v] < k && !del[v]) {
        del[v] = 1;
        q.push(v);
        cnt--;
      }
    }
  }
}
int main() {
  memset(deg, 0, sizeof deg);
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", x + i, y + i);
    deg[x[i]]++;
    deg[y[i]]++;
    G[x[i]].insert(y[i]);
    G[y[i]].insert(x[i]);
  }
  cnt = n;
  for (int i = 1; i <= n; i++) Del(i);
  for (int i = m; i; i--) {
    ans[i] = cnt;
    if (!del[x[i]]) deg[y[i]]--;
    if (!del[y[i]]) deg[x[i]]--;
    G[x[i]].erase(y[i]);
    G[y[i]].erase(x[i]);
    Del(x[i]);
    Del(y[i]);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 2e5 + 3;
int n, m, k, deg[N], x[N], y[N], ans[N], cnt;
bool del[N];
set<int> G[N];
set<int>::iterator it;
void Del(int rt) {
  if (deg[rt] >= k || del[rt]) return;
  queue<int> q;
  q.push(rt);
  del[rt] = 1;
  cnt--;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (it = G[u].begin(); it != G[u].end(); it++) {
      int v = *it;
      deg[v]--;
      if (deg[v] < k && !del[v]) {
        del[v] = 1;
        q.push(v);
        cnt--;
      }
    }
  }
}
int main() {
  memset(deg, 0, sizeof deg);
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", x + i, y + i);
    deg[x[i]]++;
    deg[y[i]]++;
    G[x[i]].insert(y[i]);
    G[y[i]].insert(x[i]);
  }
  cnt = n;
  for (int i = 1; i <= n; i++) Del(i);
  for (int i = m; i; i--) {
    ans[i] = cnt;
    if (!del[x[i]]) deg[y[i]]--;
    if (!del[y[i]]) deg[x[i]]--;
    G[x[i]].erase(y[i]);
    G[y[i]].erase(x[i]);
    Del(x[i]);
    Del(y[i]);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```