#include <bits/stdc++.h>
const int maxn = 2e5 + 3;
int n, m, k, rest, eu[maxn], ev[maxn], ans[maxn];
std::set<int> G[maxn];
bool poped[maxn];
std::queue<int> q;
void delEdge(int u, int v) {
  G[u].erase(v);
  G[v].erase(u);
  if (G[u].size() < k && !poped[u]) {
    poped[u] = true;
    --rest;
    q.push(u);
  }
  if (G[v].size() < k && !poped[v]) {
    poped[v] = true;
    --rest;
    q.push(v);
  }
}
void erase() {
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    std::set<int> E = G[u];
    for (const auto v : E) {
      delEdge(u, v);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", eu + i, ev + i);
    G[eu[i]].insert(ev[i]);
    G[ev[i]].insert(eu[i]);
  }
  rest = n;
  for (int i = 1; i <= n; ++i) {
    if (G[i].size() < k) {
      poped[i] = true;
      --rest;
      q.push(i);
    }
  }
  erase();
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = rest;
    delEdge(eu[i], ev[i]);
    erase();
  }
  for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
  return 0;
}

### 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>
const int maxn = 2e5 + 3;
int n, m, k, rest, eu[maxn], ev[maxn], ans[maxn];
std::set<int> G[maxn];
bool poped[maxn];
std::queue<int> q;
void delEdge(int u, int v) {
  G[u].erase(v);
  G[v].erase(u);
  if (G[u].size() < k && !poped[u]) {
    poped[u] = true;
    --rest;
    q.push(u);
  }
  if (G[v].size() < k && !poped[v]) {
    poped[v] = true;
    --rest;
    q.push(v);
  }
}
void erase() {
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    std::set<int> E = G[u];
    for (const auto v : E) {
      delEdge(u, v);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", eu + i, ev + i);
    G[eu[i]].insert(ev[i]);
    G[ev[i]].insert(eu[i]);
  }
  rest = n;
  for (int i = 1; i <= n; ++i) {
    if (G[i].size() < k) {
      poped[i] = true;
      --rest;
      q.push(i);
    }
  }
  erase();
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = rest;
    delEdge(eu[i], ev[i]);
    erase();
  }
  for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
  return 0;
}

```