#include <bits/stdc++.h>
template <class T>
using MinHeap = std::priority_queue<T, std::vector<T>, std::greater<T>>;
const int MAX_N = 2e5 + 7;
struct bg_edge {
  int a, b;
} edges[MAX_N];
struct bg_state {
  int node;
  long unsigned int deg;
  bool operator>(const bg_state& o) const { return deg > o.deg; }
};
static inline void setio(void);
int n, m;
long unsigned int k;
std::set<int> gr[MAX_N];
int ans[MAX_N], in[MAX_N];
int main(void) {
  setio();
  memset(in, 1, sizeof(in));
  std::cin >> n >> m >> k;
  for (int i = 0, a, b; i < m; ++i) {
    std::cin >> a >> b;
    --a;
    --b;
    gr[a].insert(b);
    gr[b].insert(a);
    edges[i] = {a, b};
  }
  MinHeap<bg_state> pq;
  for (int i = 0; i < n; ++i) pq.push(bg_state{i, gr[i].size()});
  std::function<int(void)> doit = [&](void) {
    int rem = 0;
    while (pq.size() && pq.top().deg < k) {
      bg_state now = pq.top();
      pq.pop();
      if (now.deg != gr[now.node].size() || !in[now.node]) continue;
      ++rem;
      in[now.node] = false;
      for (const auto& nei : gr[now.node]) {
        gr[nei].erase(now.node);
        pq.push(bg_state{nei, gr[nei].size()});
      }
    }
    return rem;
  };
  ans[m] = n;
  for (int i = m; i > 0; --i) {
    if (i < m) {
      gr[edges[i].a].erase(edges[i].b);
      gr[edges[i].b].erase(edges[i].a);
      pq.push(bg_state{edges[i].a, gr[edges[i].a].size()});
      pq.push(bg_state{edges[i].b, gr[edges[i].b].size()});
    }
    ans[i - 1] = ans[i] - doit();
  }
  for (int i = 0; i < m; ++i) std::cout << ans[i] << "\n";
  return 0;
}
static inline void setio(void) {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.precision(10);
  std::cout << std::fixed;
}

### Prompt

Please provide a cpp coded solution 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>
template <class T>
using MinHeap = std::priority_queue<T, std::vector<T>, std::greater<T>>;
const int MAX_N = 2e5 + 7;
struct bg_edge {
  int a, b;
} edges[MAX_N];
struct bg_state {
  int node;
  long unsigned int deg;
  bool operator>(const bg_state& o) const { return deg > o.deg; }
};
static inline void setio(void);
int n, m;
long unsigned int k;
std::set<int> gr[MAX_N];
int ans[MAX_N], in[MAX_N];
int main(void) {
  setio();
  memset(in, 1, sizeof(in));
  std::cin >> n >> m >> k;
  for (int i = 0, a, b; i < m; ++i) {
    std::cin >> a >> b;
    --a;
    --b;
    gr[a].insert(b);
    gr[b].insert(a);
    edges[i] = {a, b};
  }
  MinHeap<bg_state> pq;
  for (int i = 0; i < n; ++i) pq.push(bg_state{i, gr[i].size()});
  std::function<int(void)> doit = [&](void) {
    int rem = 0;
    while (pq.size() && pq.top().deg < k) {
      bg_state now = pq.top();
      pq.pop();
      if (now.deg != gr[now.node].size() || !in[now.node]) continue;
      ++rem;
      in[now.node] = false;
      for (const auto& nei : gr[now.node]) {
        gr[nei].erase(now.node);
        pq.push(bg_state{nei, gr[nei].size()});
      }
    }
    return rem;
  };
  ans[m] = n;
  for (int i = m; i > 0; --i) {
    if (i < m) {
      gr[edges[i].a].erase(edges[i].b);
      gr[edges[i].b].erase(edges[i].a);
      pq.push(bg_state{edges[i].a, gr[edges[i].a].size()});
      pq.push(bg_state{edges[i].b, gr[edges[i].b].size()});
    }
    ans[i - 1] = ans[i] - doit();
  }
  for (int i = 0; i < m; ++i) std::cout << ans[i] << "\n";
  return 0;
}
static inline void setio(void) {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.precision(10);
  std::cout << std::fixed;
}

```