#include <bits/stdc++.h>
using namespace std;
int Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%lld", &x);
  return x;
}
double Double() {
  double x;
  scanf("%lf", &x);
  return x;
}
const int N = 2 * (int)1e5 + 5;
const long long MOD = (int)1e9 + 7;
int us[N], vs[N], ans[N], degree[N];
bool can[N];
set<pair<int, int> > trip;
vector<pair<int, int> > g[N];
int main() {
  int n = Int(), m = Int(), k = Int();
  memset(can, true, sizeof can);
  for (int i = 1; i <= m; i++) {
    us[i] = Int(), vs[i] = Int();
    g[us[i]].push_back({vs[i], i});
    g[vs[i]].push_back({us[i], i});
    degree[us[i]]++;
    degree[vs[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    trip.insert({degree[i], i});
  }
  while (!trip.empty() and trip.begin()->first < k) {
    int u = trip.begin()->second;
    for (auto j : g[u]) {
      if (can[j.first]) {
        trip.erase({degree[j.first], j.first});
        degree[j.first]--;
        trip.insert({degree[j.first], j.first});
      }
    }
    trip.erase({degree[u], u});
    can[u] = false;
  }
  for (int i = m; i > 0; i--) {
    ans[i] = trip.size();
    if (can[us[i]] and can[vs[i]]) {
      trip.erase({degree[us[i]], us[i]});
      degree[us[i]]--;
      trip.insert({degree[us[i]], us[i]});
      trip.erase({degree[vs[i]], vs[i]});
      degree[vs[i]]--;
      trip.insert({degree[vs[i]], vs[i]});
      while (!trip.empty() and trip.begin()->first < k) {
        int u = trip.begin()->second;
        for (auto j : g[u]) {
          if (j.second >= i) continue;
          if (can[j.first]) {
            trip.erase({degree[j.first], j.first});
            degree[j.first]--;
            trip.insert({degree[j.first], j.first});
          }
        }
        trip.erase({degree[u], u});
        can[u] = false;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
  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;
int Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%lld", &x);
  return x;
}
double Double() {
  double x;
  scanf("%lf", &x);
  return x;
}
const int N = 2 * (int)1e5 + 5;
const long long MOD = (int)1e9 + 7;
int us[N], vs[N], ans[N], degree[N];
bool can[N];
set<pair<int, int> > trip;
vector<pair<int, int> > g[N];
int main() {
  int n = Int(), m = Int(), k = Int();
  memset(can, true, sizeof can);
  for (int i = 1; i <= m; i++) {
    us[i] = Int(), vs[i] = Int();
    g[us[i]].push_back({vs[i], i});
    g[vs[i]].push_back({us[i], i});
    degree[us[i]]++;
    degree[vs[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    trip.insert({degree[i], i});
  }
  while (!trip.empty() and trip.begin()->first < k) {
    int u = trip.begin()->second;
    for (auto j : g[u]) {
      if (can[j.first]) {
        trip.erase({degree[j.first], j.first});
        degree[j.first]--;
        trip.insert({degree[j.first], j.first});
      }
    }
    trip.erase({degree[u], u});
    can[u] = false;
  }
  for (int i = m; i > 0; i--) {
    ans[i] = trip.size();
    if (can[us[i]] and can[vs[i]]) {
      trip.erase({degree[us[i]], us[i]});
      degree[us[i]]--;
      trip.insert({degree[us[i]], us[i]});
      trip.erase({degree[vs[i]], vs[i]});
      degree[vs[i]]--;
      trip.insert({degree[vs[i]], vs[i]});
      while (!trip.empty() and trip.begin()->first < k) {
        int u = trip.begin()->second;
        for (auto j : g[u]) {
          if (j.second >= i) continue;
          if (can[j.first]) {
            trip.erase({degree[j.first], j.first});
            degree[j.first]--;
            trip.insert({degree[j.first], j.first});
          }
        }
        trip.erase({degree[u], u});
        can[u] = false;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}

```