#include <bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr int N = 2e5 + 5;
int ans[N];
int degree[N];
int is_good[N];
vector<pair<int, int>> E[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, k;
  cin >> n >> m >> k;
  vector<pair<int, int>> e(m);
  for (int i = 0; i < m; i++) {
    auto& [x, y] = e[i];
    cin >> x >> y;
    x--;
    y--;
    degree[x]++;
    degree[y]++;
    E[x].push_back({y, i});
    E[y].push_back({x, i});
  }
  set<pair<int, int>> good;
  for (int i = 0; i < n; i++) good.insert({degree[i], i});
  fill_n(is_good, n, 1);
  auto keep_good = [&](int t) -> void {
    while (!good.empty() && good.begin()->first < k) {
      int x = good.begin()->second;
      good.erase(good.begin());
      for (const auto& [y, i] : E[x]) {
        if (is_good[y] && i < t) {
          good.erase({degree[y], y});
          degree[y]--;
          good.insert({degree[y], y});
        }
      }
      is_good[x] = 0;
    }
  };
  keep_good(m);
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = good.size();
    const auto& [x, y] = e[i];
    if (is_good[x] && is_good[y]) {
      good.erase({degree[x], x});
      degree[x]--;
      good.insert({degree[x], x});
      good.erase({degree[y], y});
      degree[y]--;
      good.insert({degree[y], y});
      keep_good(i);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << "\n";
}

### 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;
using LL = long long;
constexpr int N = 2e5 + 5;
int ans[N];
int degree[N];
int is_good[N];
vector<pair<int, int>> E[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, k;
  cin >> n >> m >> k;
  vector<pair<int, int>> e(m);
  for (int i = 0; i < m; i++) {
    auto& [x, y] = e[i];
    cin >> x >> y;
    x--;
    y--;
    degree[x]++;
    degree[y]++;
    E[x].push_back({y, i});
    E[y].push_back({x, i});
  }
  set<pair<int, int>> good;
  for (int i = 0; i < n; i++) good.insert({degree[i], i});
  fill_n(is_good, n, 1);
  auto keep_good = [&](int t) -> void {
    while (!good.empty() && good.begin()->first < k) {
      int x = good.begin()->second;
      good.erase(good.begin());
      for (const auto& [y, i] : E[x]) {
        if (is_good[y] && i < t) {
          good.erase({degree[y], y});
          degree[y]--;
          good.insert({degree[y], y});
        }
      }
      is_good[x] = 0;
    }
  };
  keep_good(m);
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = good.size();
    const auto& [x, y] = e[i];
    if (is_good[x] && is_good[y]) {
      good.erase({degree[x], x});
      degree[x]--;
      good.insert({degree[x], x});
      good.erase({degree[y], y});
      degree[y]--;
      good.insert({degree[y], y});
      keep_good(i);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << "\n";
}

```