#include <bits/stdc++.h>
using namespace std;
const int inf32 = 1e9 + 9;
const long long inf64 = 1e18 + 18;
const int N = 2e5 + 5;
const long long mod = 1e9 + 9;
vector<int> gr[N];
set<pair<int, int> > deg;
int pw[N];
bool used[N];
void solve() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  vector<pair<int, int> > edges(m);
  vector<int> ans(m);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &edges[i].first, &edges[i].second);
    --edges[i].first, --edges[i].second;
    gr[edges[i].first].push_back(edges[i].second);
    gr[edges[i].second].push_back(edges[i].first);
  }
  for (int i = 0; i < n; ++i) {
    deg.insert({gr[i].size(), i});
    pw[i] = gr[i].size();
  }
  memset(used, true, N);
  set<pair<int, int> > usedP;
  for (int i = m - 1; i >= 0; --i) {
    while (!deg.empty() && deg.begin()->first < k) {
      pair<int, int> p = *deg.begin();
      used[deg.begin()->second] = false;
      deg.erase(deg.begin());
      for (auto &x : gr[p.second]) {
        if (used[x] &&
            !(usedP.count({x, p.second}) || usedP.count({p.second, x}))) {
          deg.erase({pw[x], x});
          deg.insert({--pw[x], x});
        }
      }
    }
    ans[i] = deg.size();
    if (!ans[i]) break;
    if (used[edges[i].first] && used[edges[i].second]) {
      deg.erase({pw[edges[i].first], edges[i].first});
      deg.insert({--pw[edges[i].first], edges[i].first});
      deg.erase({pw[edges[i].second], edges[i].second});
      deg.insert({--pw[edges[i].second], edges[i].second});
      usedP.insert({edges[i].first, edges[i].second});
    }
  }
  for (auto &x : ans) printf("%d\n", x);
}
int main() {
  solve();
  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>
using namespace std;
const int inf32 = 1e9 + 9;
const long long inf64 = 1e18 + 18;
const int N = 2e5 + 5;
const long long mod = 1e9 + 9;
vector<int> gr[N];
set<pair<int, int> > deg;
int pw[N];
bool used[N];
void solve() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  vector<pair<int, int> > edges(m);
  vector<int> ans(m);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &edges[i].first, &edges[i].second);
    --edges[i].first, --edges[i].second;
    gr[edges[i].first].push_back(edges[i].second);
    gr[edges[i].second].push_back(edges[i].first);
  }
  for (int i = 0; i < n; ++i) {
    deg.insert({gr[i].size(), i});
    pw[i] = gr[i].size();
  }
  memset(used, true, N);
  set<pair<int, int> > usedP;
  for (int i = m - 1; i >= 0; --i) {
    while (!deg.empty() && deg.begin()->first < k) {
      pair<int, int> p = *deg.begin();
      used[deg.begin()->second] = false;
      deg.erase(deg.begin());
      for (auto &x : gr[p.second]) {
        if (used[x] &&
            !(usedP.count({x, p.second}) || usedP.count({p.second, x}))) {
          deg.erase({pw[x], x});
          deg.insert({--pw[x], x});
        }
      }
    }
    ans[i] = deg.size();
    if (!ans[i]) break;
    if (used[edges[i].first] && used[edges[i].second]) {
      deg.erase({pw[edges[i].first], edges[i].first});
      deg.insert({--pw[edges[i].first], edges[i].first});
      deg.erase({pw[edges[i].second], edges[i].second});
      deg.insert({--pw[edges[i].second], edges[i].second});
      usedP.insert({edges[i].first, edges[i].second});
    }
  }
  for (auto &x : ans) printf("%d\n", x);
}
int main() {
  solve();
  return 0;
}

```