#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int sum = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
  return sum * ff;
}
const int mod = 1e9 + 7;
const int mo = 998244353;
const int N = 2e5 + 5;
int n, m, s, du[N], ok[N], ux[N], uy[N], sum[N], cnt, ans;
vector<int> G[N];
inline void del(int x) {
  queue<int> q;
  q.push(x);
  ans--;
  ok[x] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = (0); i <= ((int)G[u].size() - 1); i++) {
      int v = G[u][i];
      if (ok[v]) continue;
      if ((--du[v]) < s) q.push(v), ok[v] = 1, ans--;
    }
  }
}
int main() {
  n = read();
  m = read();
  s = read();
  for (int i = (1); i <= (m); i++) {
    int x, y;
    x = read(), y = read();
    G[x].push_back(y);
    G[y].push_back(x);
    du[y]++, du[x]++;
    ux[i] = x, uy[i] = y;
  }
  ans = n;
  for (int i = (1); i <= (n); i++)
    if (!ok[i] && du[i] < s) del(i);
  sum[m] = ans;
  for (int i = (m); i >= (1); i--) {
    int x = ux[i], y = uy[i];
    if (!ok[x]) du[y]--;
    if (!ok[y]) du[x]--;
    G[x].pop_back();
    G[y].pop_back();
    if (!ok[x] && du[x] < s) del(x);
    if (!ok[y] && du[y] < s) del(y);
    sum[i - 1] = ans;
  }
  for (int i = (1); i <= (m); i++) printf("%d\n", sum[i]);
  return 0;
}

### Prompt

Develop a solution in cpp 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;
inline int read() {
  int sum = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
  return sum * ff;
}
const int mod = 1e9 + 7;
const int mo = 998244353;
const int N = 2e5 + 5;
int n, m, s, du[N], ok[N], ux[N], uy[N], sum[N], cnt, ans;
vector<int> G[N];
inline void del(int x) {
  queue<int> q;
  q.push(x);
  ans--;
  ok[x] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = (0); i <= ((int)G[u].size() - 1); i++) {
      int v = G[u][i];
      if (ok[v]) continue;
      if ((--du[v]) < s) q.push(v), ok[v] = 1, ans--;
    }
  }
}
int main() {
  n = read();
  m = read();
  s = read();
  for (int i = (1); i <= (m); i++) {
    int x, y;
    x = read(), y = read();
    G[x].push_back(y);
    G[y].push_back(x);
    du[y]++, du[x]++;
    ux[i] = x, uy[i] = y;
  }
  ans = n;
  for (int i = (1); i <= (n); i++)
    if (!ok[i] && du[i] < s) del(i);
  sum[m] = ans;
  for (int i = (m); i >= (1); i--) {
    int x = ux[i], y = uy[i];
    if (!ok[x]) du[y]--;
    if (!ok[y]) du[x]--;
    G[x].pop_back();
    G[y].pop_back();
    if (!ok[x] && du[x] < s) del(x);
    if (!ok[y] && du[y] < s) del(y);
    sum[i - 1] = ans;
  }
  for (int i = (1); i <= (m); i++) printf("%d\n", sum[i]);
  return 0;
}

```