#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\n";
  err(++it, args...);
}
vector<pair<int, int> > edge(200003);
vector<vector<pair<int, int> > > ad(200003);
int sz[200003];
stack<int> pq;
int node;
int process(int k, int j) {
  int i, x, a, s;
  while (!pq.empty()) {
    s = pq.top();
    pq.pop();
    node--;
    sz[s] = 0;
    x = ad[s].size();
    for (i = 0; i < x; i++) {
      a = ad[s][i].first;
      if (ad[s][i].second <= j) {
        sz[a]--;
        if (sz[a] == k - 1) pq.push(a);
      }
    }
  }
  return node;
}
int main() {
  int n, m, k, i, a, b;
  scanf("%d%d%d", &n, &m, &k);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    ad[a].emplace_back(b, i);
    ad[b].emplace_back(a, i);
    sz[a]++, sz[b]++;
    edge[i] = {a, b};
  }
  vector<int> ans(m);
  for (i = 0; i < n; i++)
    if (sz[i] < k) pq.push(i);
  node = n;
  int j;
  for (i = m - 1; i >= 0; i--) {
    ans[i] = process(k, i);
    tie(a, b) = edge[i];
    if (sz[a] > 0 && sz[b] > 0) {
      if (sz[a] == k) pq.push(a);
      if (sz[b] == k) pq.push(b);
      sz[a]--, sz[b]--;
    }
  }
  for (i = 0; i < m; i++) printf("%d\n", ans[i]);
}

### Prompt

Generate 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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\n";
  err(++it, args...);
}
vector<pair<int, int> > edge(200003);
vector<vector<pair<int, int> > > ad(200003);
int sz[200003];
stack<int> pq;
int node;
int process(int k, int j) {
  int i, x, a, s;
  while (!pq.empty()) {
    s = pq.top();
    pq.pop();
    node--;
    sz[s] = 0;
    x = ad[s].size();
    for (i = 0; i < x; i++) {
      a = ad[s][i].first;
      if (ad[s][i].second <= j) {
        sz[a]--;
        if (sz[a] == k - 1) pq.push(a);
      }
    }
  }
  return node;
}
int main() {
  int n, m, k, i, a, b;
  scanf("%d%d%d", &n, &m, &k);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    ad[a].emplace_back(b, i);
    ad[b].emplace_back(a, i);
    sz[a]++, sz[b]++;
    edge[i] = {a, b};
  }
  vector<int> ans(m);
  for (i = 0; i < n; i++)
    if (sz[i] < k) pq.push(i);
  node = n;
  int j;
  for (i = m - 1; i >= 0; i--) {
    ans[i] = process(k, i);
    tie(a, b) = edge[i];
    if (sz[a] > 0 && sz[b] > 0) {
      if (sz[a] == k) pq.push(a);
      if (sz[b] == k) pq.push(b);
      sz[a]--, sz[b]--;
    }
  }
  for (i = 0; i < m; i++) printf("%d\n", ans[i]);
}

```