#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int MAX = 200005;
int ans[MAX];
vector<int> P[MAX];
pair<int, int> edges[MAX];
bool O[MAX];
int SZ[MAX];
set<pair<int, int> > wziete;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m, k;
  cin >> n >> m >> k;
  int akt = 0;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    SZ[a]++;
    SZ[b]++;
    edges[i] = make_pair(a, b);
    P[a].push_back(b);
    P[b].push_back(a);
  }
  queue<int> Q;
  for (int i = 1; i <= n; i++) {
    if (O[i]) continue;
    if (P[i].size() < k) {
      O[i] = true;
      Q.push(i);
    }
  }
  while (!Q.empty()) {
    int aktuell = Q.front();
    Q.pop();
    for (auto it : P[aktuell]) {
      int l1 = it;
      int l2 = aktuell;
      if (l1 > l2) swap(l1, l2);
      if (wziete.find(make_pair(l1, l2)) != wziete.end()) continue;
      if (!O[it] && SZ[it] - 1 < k) {
        O[it] = true;
        Q.push(it);
      }
      SZ[it]--;
      wziete.insert(make_pair(l1, l2));
    }
  }
  for (int i = 1; i <= n; i++)
    if (!O[i]) akt++;
  ans[m] = akt;
  for (int i = m; i >= 2; i--) {
    int a = edges[i].first;
    int b = edges[i].second;
    if (a > b) swap(a, b);
    if (wziete.find(make_pair(a, b)) != wziete.end()) {
      ans[i - 1] = akt;
      continue;
    }
    wziete.insert(make_pair(a, b));
    SZ[a]--;
    SZ[b]--;
    if (!O[a] && SZ[a] < k) {
      O[a] = true;
      Q.push(a);
      akt--;
    }
    if (!O[b] && SZ[b] < k) {
      O[b] = true;
      Q.push(b);
      akt--;
    }
    while (!Q.empty()) {
      int aktuell = Q.front();
      Q.pop();
      for (auto it : P[aktuell]) {
        int l1 = it;
        int l2 = aktuell;
        if (l1 > l2) swap(l1, l2);
        if (wziete.find(make_pair(l1, l2)) != wziete.end()) continue;
        if (!O[it] && SZ[it] - 1 < k) {
          O[it] = true;
          Q.push(it);
          akt--;
        }
        SZ[it]--;
        wziete.insert(make_pair(l1, l2));
      }
    }
    ans[i - 1] = akt;
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
  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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int MAX = 200005;
int ans[MAX];
vector<int> P[MAX];
pair<int, int> edges[MAX];
bool O[MAX];
int SZ[MAX];
set<pair<int, int> > wziete;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m, k;
  cin >> n >> m >> k;
  int akt = 0;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    SZ[a]++;
    SZ[b]++;
    edges[i] = make_pair(a, b);
    P[a].push_back(b);
    P[b].push_back(a);
  }
  queue<int> Q;
  for (int i = 1; i <= n; i++) {
    if (O[i]) continue;
    if (P[i].size() < k) {
      O[i] = true;
      Q.push(i);
    }
  }
  while (!Q.empty()) {
    int aktuell = Q.front();
    Q.pop();
    for (auto it : P[aktuell]) {
      int l1 = it;
      int l2 = aktuell;
      if (l1 > l2) swap(l1, l2);
      if (wziete.find(make_pair(l1, l2)) != wziete.end()) continue;
      if (!O[it] && SZ[it] - 1 < k) {
        O[it] = true;
        Q.push(it);
      }
      SZ[it]--;
      wziete.insert(make_pair(l1, l2));
    }
  }
  for (int i = 1; i <= n; i++)
    if (!O[i]) akt++;
  ans[m] = akt;
  for (int i = m; i >= 2; i--) {
    int a = edges[i].first;
    int b = edges[i].second;
    if (a > b) swap(a, b);
    if (wziete.find(make_pair(a, b)) != wziete.end()) {
      ans[i - 1] = akt;
      continue;
    }
    wziete.insert(make_pair(a, b));
    SZ[a]--;
    SZ[b]--;
    if (!O[a] && SZ[a] < k) {
      O[a] = true;
      Q.push(a);
      akt--;
    }
    if (!O[b] && SZ[b] < k) {
      O[b] = true;
      Q.push(b);
      akt--;
    }
    while (!Q.empty()) {
      int aktuell = Q.front();
      Q.pop();
      for (auto it : P[aktuell]) {
        int l1 = it;
        int l2 = aktuell;
        if (l1 > l2) swap(l1, l2);
        if (wziete.find(make_pair(l1, l2)) != wziete.end()) continue;
        if (!O[it] && SZ[it] - 1 < k) {
          O[it] = true;
          Q.push(it);
          akt--;
        }
        SZ[it]--;
        wziete.insert(make_pair(l1, l2));
      }
    }
    ans[i - 1] = akt;
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
  return 0;
}

```