#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 1;
int n, m, k, ans;
int du[MAX], X[MAX], Y[MAX], Ans[MAX];
vector<int> ve[MAX];
bool use[MAX];
void work(int x) {
  queue<int> qu;
  qu.push(x), use[x] = 1, --ans;
  while (!qu.empty()) {
    int tt = qu.front();
    qu.pop();
    for (int i = 0; i < ve[tt].size(); ++i) {
      int y = ve[tt][i];
      if (use[y]) continue;
      --du[y];
      if (du[y] < k) --ans, use[y] = 1, qu.push(y);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  ans = n;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &X[i], &Y[i]);
    ++du[X[i]], ++du[Y[i]], ve[X[i]].push_back(Y[i]), ve[Y[i]].push_back(X[i]);
  }
  for (int i = 1; i <= n; ++i)
    if (du[i] < k && !use[i]) work(i);
  Ans[m] = ans;
  for (int i = m; i >= 1; --i) {
    if (!use[Y[i]]) --du[X[i]];
    if (!use[X[i]]) --du[Y[i]];
    ve[X[i]].pop_back(), ve[Y[i]].pop_back();
    if (du[X[i]] < k && !use[X[i]]) work(X[i]);
    if (du[Y[i]] < k && !use[Y[i]]) work(Y[i]);
    Ans[i - 1] = ans;
  }
  for (int i = 1; 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>
using namespace std;
const int MAX = 2e5 + 1;
int n, m, k, ans;
int du[MAX], X[MAX], Y[MAX], Ans[MAX];
vector<int> ve[MAX];
bool use[MAX];
void work(int x) {
  queue<int> qu;
  qu.push(x), use[x] = 1, --ans;
  while (!qu.empty()) {
    int tt = qu.front();
    qu.pop();
    for (int i = 0; i < ve[tt].size(); ++i) {
      int y = ve[tt][i];
      if (use[y]) continue;
      --du[y];
      if (du[y] < k) --ans, use[y] = 1, qu.push(y);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  ans = n;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &X[i], &Y[i]);
    ++du[X[i]], ++du[Y[i]], ve[X[i]].push_back(Y[i]), ve[Y[i]].push_back(X[i]);
  }
  for (int i = 1; i <= n; ++i)
    if (du[i] < k && !use[i]) work(i);
  Ans[m] = ans;
  for (int i = m; i >= 1; --i) {
    if (!use[Y[i]]) --du[X[i]];
    if (!use[X[i]]) --du[Y[i]];
    ve[X[i]].pop_back(), ve[Y[i]].pop_back();
    if (du[X[i]] < k && !use[X[i]]) work(X[i]);
    if (du[Y[i]] < k && !use[Y[i]]) work(Y[i]);
    Ans[i - 1] = ans;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", Ans[i]);
  return 0;
}

```