#include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
bool Finish_read;
template <class T>
inline void read(T &x) {
  Finish_read = 0;
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    if (ch == EOF) return;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
  Finish_read = 1;
}
template <class T>
inline void print(T x) {
  if (x / 10 != 0) print(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void writeln(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
  putchar('\n');
}
template <class T>
inline void write(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
}
const int maxn = 200005;
set<int> G[maxn];
queue<int> q;
int n, m, k, u[maxn], v[maxn], d[maxn], vis[maxn];
vector<int> res;
inline void calc() {
  int tmp = n;
  for (int i = m; i; --i) {
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      --tmp;
      for (int y : G[x]) {
        if (G[y].find(x) == G[y].end()) continue;
        G[y].erase(x);
        d[y]--;
        if (d[y] < k && !vis[y]) q.push(y), vis[y] = 1;
      }
    }
    res.push_back(tmp);
    int x = u[i], y = v[i];
    if (G[x].find(y) == G[x].end()) goto nxt;
    G[x].erase(y);
    --d[x];
    if (d[x] < k && !vis[x]) q.push(x), vis[x] = 1;
  nxt:;
    if (G[y].find(x) == G[y].end()) continue;
    G[y].erase(x);
    --d[y];
    if (d[y] < k && !vis[y]) q.push(y), vis[y] = 1;
  }
}
inline void input() {
  read(n), read(m), read(k);
  for (int i = 1; i <= m; ++i) {
    read(u[i]), read(v[i]);
    G[u[i]].insert(v[i]);
    G[v[i]].insert(u[i]);
    d[u[i]]++, d[v[i]]++;
  }
}
int main() {
  input();
  for (int i = 1; i <= n; ++i)
    if (d[i] < k) q.push(i), vis[i] = 1;
  calc();
  for (int i = 0; i < (int)res.size(); ++i)
    printf("%d\n", res[res.size() - i - 1]);
}

### 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>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
bool Finish_read;
template <class T>
inline void read(T &x) {
  Finish_read = 0;
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    if (ch == EOF) return;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
  Finish_read = 1;
}
template <class T>
inline void print(T x) {
  if (x / 10 != 0) print(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void writeln(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
  putchar('\n');
}
template <class T>
inline void write(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
}
const int maxn = 200005;
set<int> G[maxn];
queue<int> q;
int n, m, k, u[maxn], v[maxn], d[maxn], vis[maxn];
vector<int> res;
inline void calc() {
  int tmp = n;
  for (int i = m; i; --i) {
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      --tmp;
      for (int y : G[x]) {
        if (G[y].find(x) == G[y].end()) continue;
        G[y].erase(x);
        d[y]--;
        if (d[y] < k && !vis[y]) q.push(y), vis[y] = 1;
      }
    }
    res.push_back(tmp);
    int x = u[i], y = v[i];
    if (G[x].find(y) == G[x].end()) goto nxt;
    G[x].erase(y);
    --d[x];
    if (d[x] < k && !vis[x]) q.push(x), vis[x] = 1;
  nxt:;
    if (G[y].find(x) == G[y].end()) continue;
    G[y].erase(x);
    --d[y];
    if (d[y] < k && !vis[y]) q.push(y), vis[y] = 1;
  }
}
inline void input() {
  read(n), read(m), read(k);
  for (int i = 1; i <= m; ++i) {
    read(u[i]), read(v[i]);
    G[u[i]].insert(v[i]);
    G[v[i]].insert(u[i]);
    d[u[i]]++, d[v[i]]++;
  }
}
int main() {
  input();
  for (int i = 1; i <= n; ++i)
    if (d[i] < k) q.push(i), vis[i] = 1;
  calc();
  for (int i = 0; i < (int)res.size(); ++i)
    printf("%d\n", res[res.size() - i - 1]);
}

```