#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n, m, k, sum;
int x[N], y[N];
vector<int> Edge[N];
int del[N], deg[N];
int ans[N];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
queue<int> Q;
inline void Topsort(int Start) {
  if (del[Start] || deg[Start] >= k) return;
  while (!Q.empty()) Q.pop();
  Q.push(Start);
  del[Start] = 1, --sum;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (vector<int>::iterator it = Edge[x].begin(); it != Edge[x].end();
         ++it) {
      int v = *it;
      if (del[v]) continue;
      --deg[v];
      if (deg[v] < k) del[v] = 1, --sum, Q.push(v);
    }
  }
}
int main() {
  n = read(), m = read(), k = read();
  for (int i = 1; i <= m; ++i) {
    x[i] = read(), y[i] = read();
    Edge[x[i]].push_back(y[i]), ++deg[x[i]];
    Edge[y[i]].push_back(x[i]), ++deg[y[i]];
  }
  sum = n;
  for (int i = 1; i <= n; ++i) Topsort(i);
  ans[m] = sum;
  for (int i = m; i >= 2; --i) {
    Edge[x[i]].pop_back();
    if (!del[y[i]]) --deg[x[i]];
    Edge[y[i]].pop_back();
    if (!del[x[i]]) --deg[y[i]];
    Topsort(x[i]), Topsort(y[i]);
    ans[i - 1] = sum;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 = 200005;
int n, m, k, sum;
int x[N], y[N];
vector<int> Edge[N];
int del[N], deg[N];
int ans[N];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
queue<int> Q;
inline void Topsort(int Start) {
  if (del[Start] || deg[Start] >= k) return;
  while (!Q.empty()) Q.pop();
  Q.push(Start);
  del[Start] = 1, --sum;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (vector<int>::iterator it = Edge[x].begin(); it != Edge[x].end();
         ++it) {
      int v = *it;
      if (del[v]) continue;
      --deg[v];
      if (deg[v] < k) del[v] = 1, --sum, Q.push(v);
    }
  }
}
int main() {
  n = read(), m = read(), k = read();
  for (int i = 1; i <= m; ++i) {
    x[i] = read(), y[i] = read();
    Edge[x[i]].push_back(y[i]), ++deg[x[i]];
    Edge[y[i]].push_back(x[i]), ++deg[y[i]];
  }
  sum = n;
  for (int i = 1; i <= n; ++i) Topsort(i);
  ans[m] = sum;
  for (int i = m; i >= 2; --i) {
    Edge[x[i]].pop_back();
    if (!del[y[i]]) --deg[x[i]];
    Edge[y[i]].pop_back();
    if (!del[x[i]]) --deg[y[i]];
    Topsort(x[i]), Topsort(y[i]);
    ans[i - 1] = sum;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

```