#include <bits/stdc++.h>
const int maxn = 2e5 + 10;
const int maxm = 1.5e7 + 10;
const int mod = 998244353;
using namespace std;
int n, m, k, ans;
int num[maxn], vis[maxn], sum[maxn];
pair<int, int> a[maxn];
set<int> st[maxn];
void check(int x) {
  if (num[x] >= k || !vis[x]) return;
  vis[x] = 0;
  ans--;
  queue<int> q;
  q.push(x);
  while (!q.empty()) {
    int tmp = q.front();
    q.pop();
    set<int>::iterator it = st[tmp].begin();
    while (it != st[tmp].end()) {
      num[*it]--;
      if (vis[*it] && num[*it] < k) {
        q.push(*it);
        vis[*it] = 0;
        ans--;
      }
      it++;
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  ans = n;
  for (int i = 1; i <= n; i++) vis[i] = 1;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[i].first = x, a[i].second = y;
    num[x]++, num[y]++;
    st[x].insert(y), st[y].insert(x);
  }
  for (int i = 1; i <= n; i++) check(i);
  for (int i = m - 1; i >= 0; i--) {
    sum[i] = ans;
    if (vis[a[i].second]) num[a[i].first]--;
    if (vis[a[i].first]) num[a[i].second]--;
    st[a[i].first].erase(a[i].second);
    st[a[i].second].erase(a[i].first);
    check(a[i].first), check(a[i].second);
  }
  for (int i = 0; i < m; i++) printf("%d\n", sum[i]);
  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>
const int maxn = 2e5 + 10;
const int maxm = 1.5e7 + 10;
const int mod = 998244353;
using namespace std;
int n, m, k, ans;
int num[maxn], vis[maxn], sum[maxn];
pair<int, int> a[maxn];
set<int> st[maxn];
void check(int x) {
  if (num[x] >= k || !vis[x]) return;
  vis[x] = 0;
  ans--;
  queue<int> q;
  q.push(x);
  while (!q.empty()) {
    int tmp = q.front();
    q.pop();
    set<int>::iterator it = st[tmp].begin();
    while (it != st[tmp].end()) {
      num[*it]--;
      if (vis[*it] && num[*it] < k) {
        q.push(*it);
        vis[*it] = 0;
        ans--;
      }
      it++;
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  ans = n;
  for (int i = 1; i <= n; i++) vis[i] = 1;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[i].first = x, a[i].second = y;
    num[x]++, num[y]++;
    st[x].insert(y), st[y].insert(x);
  }
  for (int i = 1; i <= n; i++) check(i);
  for (int i = m - 1; i >= 0; i--) {
    sum[i] = ans;
    if (vis[a[i].second]) num[a[i].first]--;
    if (vis[a[i].first]) num[a[i].second]--;
    st[a[i].first].erase(a[i].second);
    st[a[i].second].erase(a[i].first);
    check(a[i].first), check(a[i].second);
  }
  for (int i = 0; i < m; i++) printf("%d\n", sum[i]);
  return 0;
}

```