#include <bits/stdc++.h>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
using namespace std;
const int maxn = 2e5 + 10;
int N, M, K;
set<int> g[maxn];
set<pair<int, int> > person;
int deg[maxn];
int ans[maxn];
pair<int, int> edges[maxn];
int main() {
  scanf("%d %d %d", &N, &M, &K);
  for (int i = 1; i <= M; ++i) {
    scanf("%d %d", &edges[i].first, &edges[i].second);
    g[edges[i].first].insert(edges[i].second);
    g[edges[i].second].insert(edges[i].first);
    ++deg[edges[i].first];
    ++deg[edges[i].second];
  }
  for (int i = 1; i <= N; ++i) {
    person.insert(make_pair(deg[i], i));
  }
  for (int i = M; i >= 1; --i) {
    int x = edges[i].first, y = edges[i].second;
    while (!person.empty()) {
      auto it = person.begin();
      if (it->first >= K) {
        break;
      }
      int v = it->second;
      for (int u : g[v]) {
        person.erase(make_pair(deg[u], u));
        if (deg[u]) person.insert(make_pair(--deg[u], u));
        g[u].erase(v);
      }
      g[v].clear();
      person.erase(make_pair(deg[v], v));
      deg[v] = 0;
    }
    ans[i] = person.size();
    if (g[x].find(y) != g[x].end()) {
      person.erase(make_pair(deg[x], x));
      person.erase(make_pair(deg[y], y));
      g[y].erase(x);
      g[x].erase(y);
      if (deg[x]) person.insert(make_pair(--deg[x], x));
      if (deg[y]) person.insert(make_pair(--deg[y], y));
    }
  }
  for (int i = 1; i <= M; ++i) {
    printf("%d\n", ans[i]);
  }
}

### 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>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
using namespace std;
const int maxn = 2e5 + 10;
int N, M, K;
set<int> g[maxn];
set<pair<int, int> > person;
int deg[maxn];
int ans[maxn];
pair<int, int> edges[maxn];
int main() {
  scanf("%d %d %d", &N, &M, &K);
  for (int i = 1; i <= M; ++i) {
    scanf("%d %d", &edges[i].first, &edges[i].second);
    g[edges[i].first].insert(edges[i].second);
    g[edges[i].second].insert(edges[i].first);
    ++deg[edges[i].first];
    ++deg[edges[i].second];
  }
  for (int i = 1; i <= N; ++i) {
    person.insert(make_pair(deg[i], i));
  }
  for (int i = M; i >= 1; --i) {
    int x = edges[i].first, y = edges[i].second;
    while (!person.empty()) {
      auto it = person.begin();
      if (it->first >= K) {
        break;
      }
      int v = it->second;
      for (int u : g[v]) {
        person.erase(make_pair(deg[u], u));
        if (deg[u]) person.insert(make_pair(--deg[u], u));
        g[u].erase(v);
      }
      g[v].clear();
      person.erase(make_pair(deg[v], v));
      deg[v] = 0;
    }
    ans[i] = person.size();
    if (g[x].find(y) != g[x].end()) {
      person.erase(make_pair(deg[x], x));
      person.erase(make_pair(deg[y], y));
      g[y].erase(x);
      g[x].erase(y);
      if (deg[x]) person.insert(make_pair(--deg[x], x));
      if (deg[y]) person.insert(make_pair(--deg[y], y));
    }
  }
  for (int i = 1; i <= M; ++i) {
    printf("%d\n", ans[i]);
  }
}

```