#include <bits/stdc++.h>
using namespace std;
using lint = long long;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-16;
int N, M, K;
int edges[200010][2], degree[200010];
set<int> graph[200010];
set<tuple<int, int>> nodes;
int ans[200010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << setprecision(8) << fixed;
  cin >> N >> M >> K;
  for (int m = 0; m < M; m++) cin >> edges[m][0] >> edges[m][1];
  for (int m = 0; m < M; m++) {
    graph[edges[m][0]].insert(edges[m][1]);
    graph[edges[m][1]].insert(edges[m][0]);
    degree[edges[m][0]]++;
    degree[edges[m][1]]++;
  }
  for (int n = 1; n <= N; n++) {
    nodes.insert({degree[n], n});
  }
  for (int day = M - 1; day >= 0; day--) {
    while (nodes.size() && get<0>(*nodes.begin()) < K) {
      int cur = get<1>(*nodes.begin());
      nodes.erase(nodes.begin());
      for (const int &nxt : graph[cur]) {
        if (nodes.count({degree[nxt], nxt})) {
          nodes.erase({degree[nxt], nxt});
          nodes.insert({--degree[nxt], nxt});
        }
      }
    }
    if (nodes.empty()) break;
    ans[day] = nodes.size();
    if (nodes.count({degree[edges[day][0]], edges[day][0]}) &&
        nodes.count({degree[edges[day][1]], edges[day][1]})) {
      for (int e = 0; e < 2; e++) {
        nodes.erase({degree[edges[day][e]], edges[day][e]});
        degree[edges[day][e]]--;
        nodes.insert({degree[edges[day][e]], edges[day][e]});
        graph[edges[day][e]].erase(edges[day][1 - e]);
      }
    }
  }
  for (int ai = 0; ai < M; ai++) {
    cout << ans[ai] << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 lint = long long;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-16;
int N, M, K;
int edges[200010][2], degree[200010];
set<int> graph[200010];
set<tuple<int, int>> nodes;
int ans[200010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << setprecision(8) << fixed;
  cin >> N >> M >> K;
  for (int m = 0; m < M; m++) cin >> edges[m][0] >> edges[m][1];
  for (int m = 0; m < M; m++) {
    graph[edges[m][0]].insert(edges[m][1]);
    graph[edges[m][1]].insert(edges[m][0]);
    degree[edges[m][0]]++;
    degree[edges[m][1]]++;
  }
  for (int n = 1; n <= N; n++) {
    nodes.insert({degree[n], n});
  }
  for (int day = M - 1; day >= 0; day--) {
    while (nodes.size() && get<0>(*nodes.begin()) < K) {
      int cur = get<1>(*nodes.begin());
      nodes.erase(nodes.begin());
      for (const int &nxt : graph[cur]) {
        if (nodes.count({degree[nxt], nxt})) {
          nodes.erase({degree[nxt], nxt});
          nodes.insert({--degree[nxt], nxt});
        }
      }
    }
    if (nodes.empty()) break;
    ans[day] = nodes.size();
    if (nodes.count({degree[edges[day][0]], edges[day][0]}) &&
        nodes.count({degree[edges[day][1]], edges[day][1]})) {
      for (int e = 0; e < 2; e++) {
        nodes.erase({degree[edges[day][e]], edges[day][e]});
        degree[edges[day][e]]--;
        nodes.insert({degree[edges[day][e]], edges[day][e]});
        graph[edges[day][e]].erase(edges[day][1 - e]);
      }
    }
  }
  for (int ai = 0; ai < M; ai++) {
    cout << ans[ai] << '\n';
  }
  return 0;
}

```