#include <bits/stdc++.h>
using namespace std;
const int OO = 0x3f3f3f3f;
const double eps = (1e-10);
stringstream out;
int n, m, k, u, v, deg[200005];
bool isGood[200005];
vector<int> ans(200005);
set<pair<int, int> > good;
vector<pair<int, int> > edges(200005);
vector<vector<pair<int, int> > > adj(200005);
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cin >> n >> m >> k;
  for (int i = 0; i < (int)(m); ++i) {
    cin >> u >> v;
    --u, --v;
    edges[i].first = u, edges[i].second = v;
    adj[u].push_back(make_pair(v, i)), adj[v].push_back(make_pair(u, i));
    ++deg[u], ++deg[v];
  }
  for (int i = 0; i < (int)(n); ++i)
    good.insert(make_pair(deg[i], i)), isGood[i] = true;
  while (!good.empty() && good.begin()->first < k) {
    u = good.begin()->second;
    for (auto p : adj[u]) {
      v = p.first;
      if (isGood[v]) {
        good.erase(make_pair(deg[v], v));
        --deg[v];
        good.insert(make_pair(deg[v], v));
      }
    }
    isGood[u] = false;
    good.erase(make_pair(deg[u], u));
  }
  for (int i = (m - 1); i >= (int)(0); --i) {
    ans[i] = ((int)((good).size()));
    u = edges[i].first, v = edges[i].second;
    if (isGood[u] && isGood[v]) {
      good.erase(make_pair(deg[u], u));
      --deg[u];
      good.insert(make_pair(deg[u], u));
      good.erase(make_pair(deg[v], v));
      --deg[v];
      good.insert(make_pair(deg[v], v));
      while (!good.empty() && good.begin()->first < k) {
        u = good.begin()->second;
        for (auto p : adj[u]) {
          if (p.second >= i) continue;
          v = p.first;
          if (isGood[v]) {
            good.erase(make_pair(deg[v], v));
            --deg[v];
            good.insert(make_pair(deg[v], v));
          }
        }
        isGood[u] = false;
        good.erase(make_pair(deg[u], u));
      }
    }
  }
  for (int i = 0; i < (int)(m); ++i) out << ans[i] << '\n';
  cout << out.str();
  return 0;
}

### Prompt

In cpp, your task is to solve 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 OO = 0x3f3f3f3f;
const double eps = (1e-10);
stringstream out;
int n, m, k, u, v, deg[200005];
bool isGood[200005];
vector<int> ans(200005);
set<pair<int, int> > good;
vector<pair<int, int> > edges(200005);
vector<vector<pair<int, int> > > adj(200005);
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cin >> n >> m >> k;
  for (int i = 0; i < (int)(m); ++i) {
    cin >> u >> v;
    --u, --v;
    edges[i].first = u, edges[i].second = v;
    adj[u].push_back(make_pair(v, i)), adj[v].push_back(make_pair(u, i));
    ++deg[u], ++deg[v];
  }
  for (int i = 0; i < (int)(n); ++i)
    good.insert(make_pair(deg[i], i)), isGood[i] = true;
  while (!good.empty() && good.begin()->first < k) {
    u = good.begin()->second;
    for (auto p : adj[u]) {
      v = p.first;
      if (isGood[v]) {
        good.erase(make_pair(deg[v], v));
        --deg[v];
        good.insert(make_pair(deg[v], v));
      }
    }
    isGood[u] = false;
    good.erase(make_pair(deg[u], u));
  }
  for (int i = (m - 1); i >= (int)(0); --i) {
    ans[i] = ((int)((good).size()));
    u = edges[i].first, v = edges[i].second;
    if (isGood[u] && isGood[v]) {
      good.erase(make_pair(deg[u], u));
      --deg[u];
      good.insert(make_pair(deg[u], u));
      good.erase(make_pair(deg[v], v));
      --deg[v];
      good.insert(make_pair(deg[v], v));
      while (!good.empty() && good.begin()->first < k) {
        u = good.begin()->second;
        for (auto p : adj[u]) {
          if (p.second >= i) continue;
          v = p.first;
          if (isGood[v]) {
            good.erase(make_pair(deg[v], v));
            --deg[v];
            good.insert(make_pair(deg[v], v));
          }
        }
        isGood[u] = false;
        good.erase(make_pair(deg[u], u));
      }
    }
  }
  for (int i = 0; i < (int)(m); ++i) out << ans[i] << '\n';
  cout << out.str();
  return 0;
}

```