#include <bits/stdc++.h>
using namespace std;
const int maxn = 200100;
struct node {
  int v, u;
} e[maxn + 10];
set<int> s[maxn];
int k, ans, in[maxn], vis[maxn], ret[maxn], n, m;
queue<int> q;
void dele(int x) {
  set<int>::iterator iter;
  int u, v;
  if (vis[x] || in[x] >= k) return;
  vis[x] = 1;
  q.push(x);
  --ans;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (iter = s[u].begin(); iter != s[u].end(); iter++) {
      v = *iter;
      in[v]--;
      if (in[v] < k && !vis[v]) {
        vis[v] = 1;
        q.push(v);
        --ans;
      }
    }
  }
}
int main() {
  int i;
  scanf("%d%d%d", &n, &m, &k);
  memset(in, 0, sizeof(in));
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &e[i].u, &e[i].v);
    s[e[i].u].insert(e[i].v);
    s[e[i].v].insert(e[i].u);
    in[e[i].u]++;
    in[e[i].v]++;
  }
  memset(vis, 0, sizeof(vis));
  for (i = 1; i <= n; i++) {
    dele(i);
  }
  ans = 0;
  for (i = 1; i <= n; i++) {
    if (!vis[i]) ans++;
  }
  ret[m] = ans;
  for (i = m; i >= 1; i--) {
    if (!vis[e[i].v]) in[e[i].u]--;
    if (!vis[e[i].u]) in[e[i].v]--;
    s[e[i].u].erase(e[i].v);
    s[e[i].v].erase(e[i].u);
    dele(e[i].u);
    dele(e[i].v);
    ret[i - 1] = ans;
  }
  for (i = 1; i <= m; i++) {
    printf("%d\n", ret[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 maxn = 200100;
struct node {
  int v, u;
} e[maxn + 10];
set<int> s[maxn];
int k, ans, in[maxn], vis[maxn], ret[maxn], n, m;
queue<int> q;
void dele(int x) {
  set<int>::iterator iter;
  int u, v;
  if (vis[x] || in[x] >= k) return;
  vis[x] = 1;
  q.push(x);
  --ans;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (iter = s[u].begin(); iter != s[u].end(); iter++) {
      v = *iter;
      in[v]--;
      if (in[v] < k && !vis[v]) {
        vis[v] = 1;
        q.push(v);
        --ans;
      }
    }
  }
}
int main() {
  int i;
  scanf("%d%d%d", &n, &m, &k);
  memset(in, 0, sizeof(in));
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &e[i].u, &e[i].v);
    s[e[i].u].insert(e[i].v);
    s[e[i].v].insert(e[i].u);
    in[e[i].u]++;
    in[e[i].v]++;
  }
  memset(vis, 0, sizeof(vis));
  for (i = 1; i <= n; i++) {
    dele(i);
  }
  ans = 0;
  for (i = 1; i <= n; i++) {
    if (!vis[i]) ans++;
  }
  ret[m] = ans;
  for (i = m; i >= 1; i--) {
    if (!vis[e[i].v]) in[e[i].u]--;
    if (!vis[e[i].u]) in[e[i].v]--;
    s[e[i].u].erase(e[i].v);
    s[e[i].v].erase(e[i].u);
    dele(e[i].u);
    dele(e[i].v);
    ret[i - 1] = ans;
  }
  for (i = 1; i <= m; i++) {
    printf("%d\n", ret[i]);
  }
  return 0;
}

```