#include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
const long long maxn = (long long)1e6 + 5;
const long long mod = (long long)998244353;
const long long inf = 0x3f3f3f3f3f3f3f3f;
pair<long long, long long> a[maxn];
long long in[maxn];
vector<long long> v[maxn];
bool vis[maxn];
long long n, m, k;
long long ans;
void f(long long x) {
  queue<long long> q;
  q.push(x);
  vis[x] = true;
  --ans;
  while (!q.empty()) {
    long long now = q.front();
    q.pop();
    for (auto &i : v[now])
      if (!vis[i] && --in[i] < k) ans--, q.push(i), vis[i] = true;
  }
}
long long fin[maxn];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> k;
  ans = n;
  for (long long i = 1; i <= m; ++i) {
    cin >> a[i].first >> a[i].second;
    ++in[a[i].first], ++in[a[i].second];
    v[a[i].first].push_back(a[i].second);
    v[a[i].second].push_back(a[i].first);
  }
  for (long long i = 1; i <= n; ++i)
    if (!vis[i] && in[i] < k) f(i);
  fin[m] = ans;
  for (long long i = m; i >= 1; --i) {
    if (!vis[a[i].first]) --in[a[i].second];
    if (!vis[a[i].second]) --in[a[i].first];
    v[a[i].first].pop_back();
    v[a[i].second].pop_back();
    if (in[a[i].first] < k && !vis[a[i].first]) f(a[i].first);
    if (in[a[i].second] < k && !vis[a[i].second]) f(a[i].second);
    fin[i - 1] = ans;
  }
  for (long long i = 1; i <= m; ++i) cout << fin[i] << '\n';
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
const long long maxn = (long long)1e6 + 5;
const long long mod = (long long)998244353;
const long long inf = 0x3f3f3f3f3f3f3f3f;
pair<long long, long long> a[maxn];
long long in[maxn];
vector<long long> v[maxn];
bool vis[maxn];
long long n, m, k;
long long ans;
void f(long long x) {
  queue<long long> q;
  q.push(x);
  vis[x] = true;
  --ans;
  while (!q.empty()) {
    long long now = q.front();
    q.pop();
    for (auto &i : v[now])
      if (!vis[i] && --in[i] < k) ans--, q.push(i), vis[i] = true;
  }
}
long long fin[maxn];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> k;
  ans = n;
  for (long long i = 1; i <= m; ++i) {
    cin >> a[i].first >> a[i].second;
    ++in[a[i].first], ++in[a[i].second];
    v[a[i].first].push_back(a[i].second);
    v[a[i].second].push_back(a[i].first);
  }
  for (long long i = 1; i <= n; ++i)
    if (!vis[i] && in[i] < k) f(i);
  fin[m] = ans;
  for (long long i = m; i >= 1; --i) {
    if (!vis[a[i].first]) --in[a[i].second];
    if (!vis[a[i].second]) --in[a[i].first];
    v[a[i].first].pop_back();
    v[a[i].second].pop_back();
    if (in[a[i].first] < k && !vis[a[i].first]) f(a[i].first);
    if (in[a[i].second] < k && !vis[a[i].second]) f(a[i].second);
    fin[i - 1] = ans;
  }
  for (long long i = 1; i <= m; ++i) cout << fin[i] << '\n';
  return 0;
}

```