#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res % mod;
}
const int N = 2e5 + 10;
int degree[N];
vector<pair<int, int> > v[N];
bool mark[N];
set<pair<int, int> > s;
int main() {
  memset(mark, 1, sizeof mark);
  int n, m, k;
  cin >> n >> m >> k;
  vector<pair<int, int> > edges(m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--, b--;
    edges[i].first = a;
    edges[i].second = b;
    degree[a]++;
    degree[b]++;
    v[a].push_back(make_pair(b, i));
    v[b].push_back(make_pair(a, i));
  }
  for (int i = 0; i < n; ++i) {
    s.insert(make_pair(degree[i], i));
  }
  while (!s.empty() && s.begin()->first < k) {
    int u = s.begin()->second;
    for (int i = 0; i < v[u].size(); ++i) {
      int ch = v[u][i].first;
      if (mark[ch]) {
        s.erase(make_pair(degree[ch], ch));
        degree[ch]--;
        s.insert(make_pair(degree[ch], ch));
      }
    }
    s.erase(make_pair(degree[u], u));
    mark[u] = 0;
  }
  vector<int> ans;
  for (int i = m - 1; i >= 0; --i) {
    ans.push_back(s.size());
    int a = edges[i].first, b = edges[i].second;
    if (mark[a] && mark[b]) {
      s.erase(make_pair(degree[a], a));
      degree[a]--;
      s.insert(make_pair(degree[a], a));
      s.erase(make_pair(degree[b], b));
      degree[b]--;
      s.insert(make_pair(degree[b], b));
      while (!s.empty() && (s.begin()->first < k)) {
        int u = s.begin()->second;
        for (int j = 0; j < v[u].size(); ++j) {
          int ch = v[u][j].first, id = v[u][j].second;
          if (id >= i) continue;
          if (mark[ch]) {
            s.erase(make_pair(degree[ch], ch));
            degree[ch]--;
            s.insert(make_pair(degree[ch], ch));
          }
        }
        s.erase(make_pair(degree[u], u));
        mark[u] = 0;
      }
    }
  }
  for (int i = ans.size() - 1; i >= 0; --i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Please formulate 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 M = 1e9 + 7;
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res % mod;
}
const int N = 2e5 + 10;
int degree[N];
vector<pair<int, int> > v[N];
bool mark[N];
set<pair<int, int> > s;
int main() {
  memset(mark, 1, sizeof mark);
  int n, m, k;
  cin >> n >> m >> k;
  vector<pair<int, int> > edges(m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--, b--;
    edges[i].first = a;
    edges[i].second = b;
    degree[a]++;
    degree[b]++;
    v[a].push_back(make_pair(b, i));
    v[b].push_back(make_pair(a, i));
  }
  for (int i = 0; i < n; ++i) {
    s.insert(make_pair(degree[i], i));
  }
  while (!s.empty() && s.begin()->first < k) {
    int u = s.begin()->second;
    for (int i = 0; i < v[u].size(); ++i) {
      int ch = v[u][i].first;
      if (mark[ch]) {
        s.erase(make_pair(degree[ch], ch));
        degree[ch]--;
        s.insert(make_pair(degree[ch], ch));
      }
    }
    s.erase(make_pair(degree[u], u));
    mark[u] = 0;
  }
  vector<int> ans;
  for (int i = m - 1; i >= 0; --i) {
    ans.push_back(s.size());
    int a = edges[i].first, b = edges[i].second;
    if (mark[a] && mark[b]) {
      s.erase(make_pair(degree[a], a));
      degree[a]--;
      s.insert(make_pair(degree[a], a));
      s.erase(make_pair(degree[b], b));
      degree[b]--;
      s.insert(make_pair(degree[b], b));
      while (!s.empty() && (s.begin()->first < k)) {
        int u = s.begin()->second;
        for (int j = 0; j < v[u].size(); ++j) {
          int ch = v[u][j].first, id = v[u][j].second;
          if (id >= i) continue;
          if (mark[ch]) {
            s.erase(make_pair(degree[ch], ch));
            degree[ch]--;
            s.insert(make_pair(degree[ch], ch));
          }
        }
        s.erase(make_pair(degree[u], u));
        mark[u] = 0;
      }
    }
  }
  for (int i = ans.size() - 1; i >= 0; --i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```