#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1123456;
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
void vout(T s) {
  cout << s << endl;
  exit(0);
}
long long bp(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a;
    a *= a;
    n >>= 1;
  }
  return res;
}
set<long long> s[MAXN];
long long n, m, k;
long long kol[MAXN];
long long ans;
bool fl[MAXN];
pair<long long, long long> a[MAXN];
queue<long long> q;
void add(long long x) {
  if (fl[x]) return;
  ans--;
  fl[x] = 1;
  q.push(x);
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  ans = n;
  vector<long long> res(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i].first >> a[i].second;
    kol[a[i].first]++;
    kol[a[i].second]++;
    s[a[i].first].insert(a[i].second);
    s[a[i].second].insert(a[i].first);
  }
  for (int i = 1; i <= n; i++)
    if (!fl[i] && kol[i] < k) {
      add(i);
      while (!q.empty()) {
        long long x = q.front();
        q.pop();
        for (auto i : s[x]) {
          kol[i]--;
          if (kol[i] < k) add(i);
        }
      }
    }
  for (int i = m - 1; i >= 0; i--) {
    res[i] = ans;
    if (!fl[a[i].first] && !fl[a[i].second]) {
      kol[a[i].first]--;
      kol[a[i].second]--;
      s[a[i].first].erase(a[i].second);
      s[a[i].second].erase(a[i].first);
      if (kol[a[i].first] < k) add(a[i].first);
      if (kol[a[i].second] < k) add(a[i].second);
      while (!q.empty()) {
        long long x = q.front();
        q.pop();
        for (auto i : s[x]) {
          kol[i]--;
          if (kol[i] < k) add(i);
        }
      }
    }
  }
  for (auto i : res) cout << i << "\n";
  return 0;
}

### 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>
using namespace std;
const long long MAXN = 1123456;
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
void vout(T s) {
  cout << s << endl;
  exit(0);
}
long long bp(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a;
    a *= a;
    n >>= 1;
  }
  return res;
}
set<long long> s[MAXN];
long long n, m, k;
long long kol[MAXN];
long long ans;
bool fl[MAXN];
pair<long long, long long> a[MAXN];
queue<long long> q;
void add(long long x) {
  if (fl[x]) return;
  ans--;
  fl[x] = 1;
  q.push(x);
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  ans = n;
  vector<long long> res(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i].first >> a[i].second;
    kol[a[i].first]++;
    kol[a[i].second]++;
    s[a[i].first].insert(a[i].second);
    s[a[i].second].insert(a[i].first);
  }
  for (int i = 1; i <= n; i++)
    if (!fl[i] && kol[i] < k) {
      add(i);
      while (!q.empty()) {
        long long x = q.front();
        q.pop();
        for (auto i : s[x]) {
          kol[i]--;
          if (kol[i] < k) add(i);
        }
      }
    }
  for (int i = m - 1; i >= 0; i--) {
    res[i] = ans;
    if (!fl[a[i].first] && !fl[a[i].second]) {
      kol[a[i].first]--;
      kol[a[i].second]--;
      s[a[i].first].erase(a[i].second);
      s[a[i].second].erase(a[i].first);
      if (kol[a[i].first] < k) add(a[i].first);
      if (kol[a[i].second] < k) add(a[i].second);
      while (!q.empty()) {
        long long x = q.front();
        q.pop();
        for (auto i : s[x]) {
          kol[i]--;
          if (kol[i] < k) add(i);
        }
      }
    }
  }
  for (auto i : res) cout << i << "\n";
  return 0;
}

```