#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
pair<int, int> edge[maxn];
int deg[maxn], ans[maxn];
bool del[maxn], vis[maxn];
set<pair<int, int> > dic;
vector<pair<int, int> > g[maxn];
int main() {
  int n, m, k, u, v;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(pair<int, int>(v, i));
    g[v].push_back(pair<int, int>(u, i));
    deg[u]++, deg[v]++;
    edge[i] = pair<int, int>(u, v);
  }
  for (int v = 1; v <= n; v++) dic.insert(pair<int, int>(deg[v], v));
  while (!dic.empty() && dic.begin()->first < k) {
    int v = dic.begin()->second;
    dic.erase(dic.begin());
    del[v] = true;
    for (int i = 0; i < (int)g[v].size(); i++) {
      int adj = g[v][i].first;
      if (del[adj]) continue;
      dic.erase(pair<int, int>(deg[adj], adj));
      dic.insert(pair<int, int>(--deg[adj], adj));
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = dic.size();
    u = edge[i].first, v = edge[i].second;
    if (!del[u] && !del[v] && !vis[i]) {
      dic.erase(pair<int, int>(deg[u], u));
      dic.insert(pair<int, int>(--deg[u], u));
      dic.erase(pair<int, int>(deg[v], v));
      dic.insert(pair<int, int>(--deg[v], v));
      vis[i] = true;
    }
    while (!dic.empty() && dic.begin()->first < k) {
      int v = dic.begin()->second;
      dic.erase(dic.begin());
      del[v] = true;
      for (int i = 0; i < (int)g[v].size(); i++) {
        int adj = g[v][i].first;
        if (del[adj] || vis[g[v][i].second]) continue;
        dic.erase(pair<int, int>(deg[adj], adj));
        dic.insert(pair<int, int>(--deg[adj], adj));
        vis[g[v][i].second] = true;
      }
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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 maxn = 200010;
pair<int, int> edge[maxn];
int deg[maxn], ans[maxn];
bool del[maxn], vis[maxn];
set<pair<int, int> > dic;
vector<pair<int, int> > g[maxn];
int main() {
  int n, m, k, u, v;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(pair<int, int>(v, i));
    g[v].push_back(pair<int, int>(u, i));
    deg[u]++, deg[v]++;
    edge[i] = pair<int, int>(u, v);
  }
  for (int v = 1; v <= n; v++) dic.insert(pair<int, int>(deg[v], v));
  while (!dic.empty() && dic.begin()->first < k) {
    int v = dic.begin()->second;
    dic.erase(dic.begin());
    del[v] = true;
    for (int i = 0; i < (int)g[v].size(); i++) {
      int adj = g[v][i].first;
      if (del[adj]) continue;
      dic.erase(pair<int, int>(deg[adj], adj));
      dic.insert(pair<int, int>(--deg[adj], adj));
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = dic.size();
    u = edge[i].first, v = edge[i].second;
    if (!del[u] && !del[v] && !vis[i]) {
      dic.erase(pair<int, int>(deg[u], u));
      dic.insert(pair<int, int>(--deg[u], u));
      dic.erase(pair<int, int>(deg[v], v));
      dic.insert(pair<int, int>(--deg[v], v));
      vis[i] = true;
    }
    while (!dic.empty() && dic.begin()->first < k) {
      int v = dic.begin()->second;
      dic.erase(dic.begin());
      del[v] = true;
      for (int i = 0; i < (int)g[v].size(); i++) {
        int adj = g[v][i].first;
        if (del[adj] || vis[g[v][i].second]) continue;
        dic.erase(pair<int, int>(deg[adj], adj));
        dic.insert(pair<int, int>(--deg[adj], adj));
        vis[g[v][i].second] = true;
      }
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

```