#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int ind[N], ans[N];
set<int> vec[N];
int U[N], V[N];
bool vis[N];
int n, m, K;
int sum = 0;
queue<int> q;
void Top() {
  while (!q.empty()) {
    ++sum;
    int u = q.front();
    q.pop();
    ind[u] = 0;
    set<int>::iterator it;
    for (it = vec[u].begin(); it != vec[u].end();) {
      int v = *it;
      if (vis[v]) {
        it++;
        continue;
      }
      vec[u].erase(it++);
      vec[v].erase(u);
      ind[v]--;
      if (!vis[v] && ind[v] < K) {
        vis[v] = 1;
        q.push(v);
      }
    }
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &K);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &U[i], &V[i]);
    int u = U[i], v = V[i];
    vec[u].insert(v);
    vec[v].insert(u);
    ind[u]++;
    ind[v]++;
  }
  for (int i = 1; i < n + 1; ++i) {
    if (ind[i] < K) {
      vis[i] = 1;
      q.push(i);
    }
  }
  Top();
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = n - sum;
    int u = U[i], v = V[i];
    if (vec[u].count(v)) ind[v]--, vec[u].erase(v);
    if (vec[v].count(u)) ind[u]--, vec[v].erase(u);
    if (!vis[v] && ind[v] < K) vis[v] = 1, q.push(v);
    if (!vis[u] && ind[u] < K) vis[u] = 1, q.push(u);
    Top();
  }
  for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please create a solution in Cpp 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 N = 2e5 + 10;
int ind[N], ans[N];
set<int> vec[N];
int U[N], V[N];
bool vis[N];
int n, m, K;
int sum = 0;
queue<int> q;
void Top() {
  while (!q.empty()) {
    ++sum;
    int u = q.front();
    q.pop();
    ind[u] = 0;
    set<int>::iterator it;
    for (it = vec[u].begin(); it != vec[u].end();) {
      int v = *it;
      if (vis[v]) {
        it++;
        continue;
      }
      vec[u].erase(it++);
      vec[v].erase(u);
      ind[v]--;
      if (!vis[v] && ind[v] < K) {
        vis[v] = 1;
        q.push(v);
      }
    }
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &K);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &U[i], &V[i]);
    int u = U[i], v = V[i];
    vec[u].insert(v);
    vec[v].insert(u);
    ind[u]++;
    ind[v]++;
  }
  for (int i = 1; i < n + 1; ++i) {
    if (ind[i] < K) {
      vis[i] = 1;
      q.push(i);
    }
  }
  Top();
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = n - sum;
    int u = U[i], v = V[i];
    if (vec[u].count(v)) ind[v]--, vec[u].erase(v);
    if (vec[v].count(u)) ind[u]--, vec[v].erase(u);
    if (!vis[v] && ind[v] < K) vis[v] = 1, q.push(v);
    if (!vis[u] && ind[u] < K) vis[u] = 1, q.push(u);
    Top();
  }
  for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
  return 0;
}

```