#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
set<pair<int, int> > sq;
set<int> to[maxn];
int n, m, k;
int a[maxn];
struct node {
  int l, r;
} q[maxn];
void update() {
  while (sq.size() > 0 && (*(sq.begin())).first < k) {
    auto x = *(sq.begin());
    sq.erase(sq.begin());
    for (auto j = to[x.second].begin(); j != to[x.second].end();) {
      auto value = *j;
      if (sq.count({a[value], value})) {
        sq.erase(sq.find({a[value], value}));
        a[value]--;
        if (a[value] > 0) sq.insert({a[value], value});
        to[x.second].erase(j++);
      } else
        j++;
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int from, t;
    cin >> from >> t;
    q[i].l = from, q[i].r = t;
    to[from].insert(t);
    to[t].insert(from);
    a[from]++, a[t]++;
  }
  for (int i = 1; i <= n; i++) sq.insert({a[i], i});
  vector<int> ans;
  for (int i = m; i >= 1; i--) {
    update();
    ans.push_back(sq.size());
    if (sq.count({a[q[i].l], q[i].l}) && sq.count({a[q[i].r], q[i].r})) {
      sq.erase(sq.find({a[q[i].l], q[i].l}));
      sq.erase(sq.find({a[q[i].r], q[i].r}));
      a[q[i].l]--, a[q[i].r]--;
      if (a[q[i].l] > 0) sq.insert({a[q[i].l], q[i].l});
      if (a[q[i].r] > 0) sq.insert({a[q[i].r], q[i].r});
      to[q[i].l].erase(q[i].r);
      to[q[i].r].erase(q[i].l);
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto &i : ans) {
    cout << i << endl;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 maxn = 2e5 + 10;
set<pair<int, int> > sq;
set<int> to[maxn];
int n, m, k;
int a[maxn];
struct node {
  int l, r;
} q[maxn];
void update() {
  while (sq.size() > 0 && (*(sq.begin())).first < k) {
    auto x = *(sq.begin());
    sq.erase(sq.begin());
    for (auto j = to[x.second].begin(); j != to[x.second].end();) {
      auto value = *j;
      if (sq.count({a[value], value})) {
        sq.erase(sq.find({a[value], value}));
        a[value]--;
        if (a[value] > 0) sq.insert({a[value], value});
        to[x.second].erase(j++);
      } else
        j++;
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int from, t;
    cin >> from >> t;
    q[i].l = from, q[i].r = t;
    to[from].insert(t);
    to[t].insert(from);
    a[from]++, a[t]++;
  }
  for (int i = 1; i <= n; i++) sq.insert({a[i], i});
  vector<int> ans;
  for (int i = m; i >= 1; i--) {
    update();
    ans.push_back(sq.size());
    if (sq.count({a[q[i].l], q[i].l}) && sq.count({a[q[i].r], q[i].r})) {
      sq.erase(sq.find({a[q[i].l], q[i].l}));
      sq.erase(sq.find({a[q[i].r], q[i].r}));
      a[q[i].l]--, a[q[i].r]--;
      if (a[q[i].l] > 0) sq.insert({a[q[i].l], q[i].l});
      if (a[q[i].r] > 0) sq.insert({a[q[i].r], q[i].r});
      to[q[i].l].erase(q[i].r);
      to[q[i].r].erase(q[i].l);
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto &i : ans) {
    cout << i << endl;
  }
  return 0;
}

```