#include <bits/stdc++.h>
using namespace std;
int l[200100], r[200100];
int n, m, k;
int de[200100];
bool ex[200100];
struct ind {
  int id, de;
  ind(int a, int b) : id(a), de(b) {}
  bool operator<(const ind &b) const {
    return de != b.de ? de < b.de : id < b.id;
  }
};
vector<int> pa[200100];
set<ind> S;
int fd(int id, int x) { return l[id] == x ? r[id] : l[id]; }
void erase(int x) {
  if (de[x] < k) return;
  S.erase(ind(x, de[x]));
  de[x]--;
  S.insert(ind(x, de[x]));
}
void dele(int x) {
  S.erase(ind(x, de[x]));
  for (auto y : pa[x])
    if (!ex[y] && de[fd(y, x)] >= k) erase(fd(y, x));
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  int a, b;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &a, &b);
    ++de[a], ++de[b];
    l[i] = a, r[i] = b;
    pa[a].push_back(i);
    pa[b].push_back(i);
  }
  for (int i = 1; i <= n; ++i) S.insert(ind(i, de[i]));
  int as[200100];
  for (int i = m; i >= 1; --i) {
    while (S.size()) {
      if ((*S.begin()).de < k)
        dele((*S.begin()).id);
      else
        break;
    }
    as[i] = S.size();
    int x = l[i], y = r[i];
    if (de[x] >= k && de[y] >= k) erase(x), erase(y);
    ex[i] = 1;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", as[i]);
  return 0;
}

### Prompt

Your task is to create 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 l[200100], r[200100];
int n, m, k;
int de[200100];
bool ex[200100];
struct ind {
  int id, de;
  ind(int a, int b) : id(a), de(b) {}
  bool operator<(const ind &b) const {
    return de != b.de ? de < b.de : id < b.id;
  }
};
vector<int> pa[200100];
set<ind> S;
int fd(int id, int x) { return l[id] == x ? r[id] : l[id]; }
void erase(int x) {
  if (de[x] < k) return;
  S.erase(ind(x, de[x]));
  de[x]--;
  S.insert(ind(x, de[x]));
}
void dele(int x) {
  S.erase(ind(x, de[x]));
  for (auto y : pa[x])
    if (!ex[y] && de[fd(y, x)] >= k) erase(fd(y, x));
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  int a, b;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &a, &b);
    ++de[a], ++de[b];
    l[i] = a, r[i] = b;
    pa[a].push_back(i);
    pa[b].push_back(i);
  }
  for (int i = 1; i <= n; ++i) S.insert(ind(i, de[i]));
  int as[200100];
  for (int i = m; i >= 1; --i) {
    while (S.size()) {
      if ((*S.begin()).de < k)
        dele((*S.begin()).id);
      else
        break;
    }
    as[i] = S.size();
    int x = l[i], y = r[i];
    if (de[x] >= k && de[y] >= k) erase(x), erase(y);
    ex[i] = 1;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", as[i]);
  return 0;
}

```