#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
vector<pair<int, int>> edges, g[N];
vector<int> ans;
set<pair<int, int>> s;
set<int> deleted;
int n, m, k, a, b, deg[N];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    edges.emplace_back(a, b);
    deg[a]++;
    deg[b]++;
    g[a].emplace_back(b, i);
    g[b].emplace_back(a, i);
  }
  for (int i = 1; i <= n; ++i) s.emplace(deg[i], i);
  for (int ind = m - 1; ind >= 0; --ind) {
    while (s.size() && s.begin()->first < k) {
      int u = s.begin()->second;
      s.erase(s.begin());
      deleted.insert(u);
      for (auto &i : g[u]) {
        int v = i.first;
        if (i.second > ind) continue;
        if (!deleted.count(v)) {
          s.erase({deg[v], v});
          --deg[v];
          s.insert({deg[v], v});
        }
      }
    }
    ans.emplace_back((int)s.size());
    int x, y;
    tie(x, y) = edges[ind];
    if (!deleted.count(x) && !deleted.count(y)) {
      s.erase({deg[x], x});
      deg[x]--;
      s.insert({deg[x], x});
      s.erase({deg[y], y});
      deg[y]--;
      s.insert({deg[y], y});
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto &i : ans) printf("%d\n", i);
  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 N = 2e5 + 10;
vector<pair<int, int>> edges, g[N];
vector<int> ans;
set<pair<int, int>> s;
set<int> deleted;
int n, m, k, a, b, deg[N];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    edges.emplace_back(a, b);
    deg[a]++;
    deg[b]++;
    g[a].emplace_back(b, i);
    g[b].emplace_back(a, i);
  }
  for (int i = 1; i <= n; ++i) s.emplace(deg[i], i);
  for (int ind = m - 1; ind >= 0; --ind) {
    while (s.size() && s.begin()->first < k) {
      int u = s.begin()->second;
      s.erase(s.begin());
      deleted.insert(u);
      for (auto &i : g[u]) {
        int v = i.first;
        if (i.second > ind) continue;
        if (!deleted.count(v)) {
          s.erase({deg[v], v});
          --deg[v];
          s.insert({deg[v], v});
        }
      }
    }
    ans.emplace_back((int)s.size());
    int x, y;
    tie(x, y) = edges[ind];
    if (!deleted.count(x) && !deleted.count(y)) {
      s.erase({deg[x], x});
      deg[x]--;
      s.insert({deg[x], x});
      s.erase({deg[y], y});
      deg[y]--;
      s.insert({deg[y], y});
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto &i : ans) printf("%d\n", i);
  return 0;
}

```