#include <bits/stdc++.h>
using namespace std;
const int maxi = 1e6 + 2;
int a[maxi], b[maxi];
int res[maxi];
vector<int> v[maxi];
string s;
int n, m, k;
int ok;
int rem[maxi];
int deg[maxi];
map<pair<int, int>, int> mp;
queue<int> q;
void skloni(int poz) {
  while (!q.empty()) {
    int x = q.front();
    deg[x] = 0;
    q.pop();
    for (int i : v[x]) {
      if (!rem[i] && deg[i] == k && !mp[{x, i}]) {
        deg[i]--;
        res[poz]--;
        q.push(i);
        rem[i] = 1;
        mp[{x, i}] = 1;
        mp[{i, x}] = 1;
      } else {
        if (!mp[{x, i}]) deg[i]--;
        mp[{x, i}] = 1;
        mp[{i, x}] = 1;
      }
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[i] = x;
    b[i] = y;
    v[x].push_back(y);
    v[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  res[m] = n;
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) {
      q.push(i);
      rem[i] = 1;
      res[m]--;
    }
  skloni(m);
  for (int i = m; i >= 1; i--) {
    res[i - 1] = res[i];
    if (!mp[{a[i], b[i]}]) {
      deg[a[i]]--;
      deg[b[i]]--;
      mp[{a[i], b[i]}] = 1;
      mp[{b[i], a[i]}] = 1;
      if (deg[a[i]] == k - 1 && !rem[a[i]]) {
        rem[a[i]] = 1;
        res[i - 1]--;
        q.push(a[i]);
      };
      if (deg[b[i]] == k - 1 && !rem[b[i]]) {
        rem[b[i]] = 1;
        res[i - 1]--;
        q.push(b[i]);
      };
      skloni(i - 1);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", res[i]);
  return 0;
}

### 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>
using namespace std;
const int maxi = 1e6 + 2;
int a[maxi], b[maxi];
int res[maxi];
vector<int> v[maxi];
string s;
int n, m, k;
int ok;
int rem[maxi];
int deg[maxi];
map<pair<int, int>, int> mp;
queue<int> q;
void skloni(int poz) {
  while (!q.empty()) {
    int x = q.front();
    deg[x] = 0;
    q.pop();
    for (int i : v[x]) {
      if (!rem[i] && deg[i] == k && !mp[{x, i}]) {
        deg[i]--;
        res[poz]--;
        q.push(i);
        rem[i] = 1;
        mp[{x, i}] = 1;
        mp[{i, x}] = 1;
      } else {
        if (!mp[{x, i}]) deg[i]--;
        mp[{x, i}] = 1;
        mp[{i, x}] = 1;
      }
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[i] = x;
    b[i] = y;
    v[x].push_back(y);
    v[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  res[m] = n;
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) {
      q.push(i);
      rem[i] = 1;
      res[m]--;
    }
  skloni(m);
  for (int i = m; i >= 1; i--) {
    res[i - 1] = res[i];
    if (!mp[{a[i], b[i]}]) {
      deg[a[i]]--;
      deg[b[i]]--;
      mp[{a[i], b[i]}] = 1;
      mp[{b[i], a[i]}] = 1;
      if (deg[a[i]] == k - 1 && !rem[a[i]]) {
        rem[a[i]] = 1;
        res[i - 1]--;
        q.push(a[i]);
      };
      if (deg[b[i]] == k - 1 && !rem[b[i]]) {
        rem[b[i]] = 1;
        res[i - 1]--;
        q.push(b[i]);
      };
      skloni(i - 1);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", res[i]);
  return 0;
}

```