#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}
template <typename T>
inline bool checkMax(T& a, const T& b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool checkMin(T& a, const T& b) {
  return a > b ? a = b, true : false;
}
const int N = 2e5 + 10;
int n, m, k, degree[N], ans[N], bl, avalib[N];
vector<pair<int, int> > G[N];
pair<int, int> f[N];
int main() {
  read(n), read(m), read(k), bl = n;
  for (register int i = 1; i <= m; ++i) {
    int u, v;
    read(u), read(v);
    ++degree[u], ++degree[v];
    G[u].push_back(make_pair(v, i));
    G[v].push_back(make_pair(u, i));
    f[i] = pair<int, int>(u, v);
  }
  queue<int> Q;
  for (register int i = 1; i <= n; ++i) {
    if (degree[i] < k) {
      --bl, Q.push(i);
    }
  }
  for (register int i = m; i; --i) {
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (auto v : G[x]) {
        if (!avalib[v.second]) {
          if (--degree[v.first] == k - 1) {
            --bl, Q.push(v.first);
          }
          avalib[v.second] = 1;
        }
      }
    }
    ans[i] = bl;
    if (!avalib[i]) {
      if (--degree[f[i].first] == k - 1) {
        --bl, Q.push(f[i].first);
      }
      if (--degree[f[i].second] == k - 1) {
        --bl, Q.push(f[i].second);
      }
      avalib[i] = 1;
    }
  }
  for (register int i = 1; i <= m; ++i) {
    printf("%d\n", ans[i]);
  }
  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;
template <typename T>
inline void read(T& x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}
template <typename T>
inline bool checkMax(T& a, const T& b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool checkMin(T& a, const T& b) {
  return a > b ? a = b, true : false;
}
const int N = 2e5 + 10;
int n, m, k, degree[N], ans[N], bl, avalib[N];
vector<pair<int, int> > G[N];
pair<int, int> f[N];
int main() {
  read(n), read(m), read(k), bl = n;
  for (register int i = 1; i <= m; ++i) {
    int u, v;
    read(u), read(v);
    ++degree[u], ++degree[v];
    G[u].push_back(make_pair(v, i));
    G[v].push_back(make_pair(u, i));
    f[i] = pair<int, int>(u, v);
  }
  queue<int> Q;
  for (register int i = 1; i <= n; ++i) {
    if (degree[i] < k) {
      --bl, Q.push(i);
    }
  }
  for (register int i = m; i; --i) {
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (auto v : G[x]) {
        if (!avalib[v.second]) {
          if (--degree[v.first] == k - 1) {
            --bl, Q.push(v.first);
          }
          avalib[v.second] = 1;
        }
      }
    }
    ans[i] = bl;
    if (!avalib[i]) {
      if (--degree[f[i].first] == k - 1) {
        --bl, Q.push(f[i].first);
      }
      if (--degree[f[i].second] == k - 1) {
        --bl, Q.push(f[i].second);
      }
      avalib[i] = 1;
    }
  }
  for (register int i = 1; i <= m; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```