#include <bits/stdc++.h>
using namespace std;
int n, m, k;
struct Node {
  int x, y;
} a[200010];
int deg[200010];
bool fl[200010];
set<int> v[200010];
int ans;
int ans1[200010];
int pos[200010];
void solve(int x) {
  if (deg[x] >= k || !fl[x]) {
    return;
  }
  queue<int> q;
  q.push(x);
  fl[x] = 0;
  ans--;
  while (!q.empty()) {
    int y = q.front();
    q.pop();
    for (auto to : v[y]) {
      deg[to]--;
      if (deg[to] < k && fl[to]) {
        fl[to] = 0;
        q.push(to);
        ans--;
      }
    }
  }
}
int main(int argc, char const *argv[]) {
  scanf("%d%d%d", &n, &m, &k);
  memset(fl, 1, sizeof(fl));
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    deg[a[i].x]++;
    deg[a[i].y]++;
    v[a[i].x].insert(a[i].y);
    v[a[i].y].insert(a[i].x);
  }
  ans = n;
  for (int i = 1; i <= n; i++) {
    solve(i);
  }
  for (int i = m; i; i--) {
    ans1[i] = ans;
    if (fl[a[i].y]) {
      deg[a[i].x]--;
    }
    if (fl[a[i].x]) {
      deg[a[i].y]--;
    }
    v[a[i].x].erase(a[i].y);
    v[a[i].y].erase(a[i].x);
    solve(a[i].x);
    solve(a[i].y);
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans1[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;
int n, m, k;
struct Node {
  int x, y;
} a[200010];
int deg[200010];
bool fl[200010];
set<int> v[200010];
int ans;
int ans1[200010];
int pos[200010];
void solve(int x) {
  if (deg[x] >= k || !fl[x]) {
    return;
  }
  queue<int> q;
  q.push(x);
  fl[x] = 0;
  ans--;
  while (!q.empty()) {
    int y = q.front();
    q.pop();
    for (auto to : v[y]) {
      deg[to]--;
      if (deg[to] < k && fl[to]) {
        fl[to] = 0;
        q.push(to);
        ans--;
      }
    }
  }
}
int main(int argc, char const *argv[]) {
  scanf("%d%d%d", &n, &m, &k);
  memset(fl, 1, sizeof(fl));
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    deg[a[i].x]++;
    deg[a[i].y]++;
    v[a[i].x].insert(a[i].y);
    v[a[i].y].insert(a[i].x);
  }
  ans = n;
  for (int i = 1; i <= n; i++) {
    solve(i);
  }
  for (int i = m; i; i--) {
    ans1[i] = ans;
    if (fl[a[i].y]) {
      deg[a[i].x]--;
    }
    if (fl[a[i].x]) {
      deg[a[i].y]--;
    }
    v[a[i].x].erase(a[i].y);
    v[a[i].y].erase(a[i].x);
    solve(a[i].x);
    solve(a[i].y);
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans1[i]);
  }
  return 0;
}

```