#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
inline void upd1(T1& a, const T2& b) {
  a = a < b ? a : b;
}
template <class T1, class T2>
inline void upd2(T1& a, const T2& b) {
  a = a > b ? a : b;
}
template <class T>
inline bool equ(const T& a, const T& b) {
  return !memcmp(&a, &b, sizeof a);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
struct ano {
  operator long long() {
    long long x = 0, y = 0, c = getchar();
    while (c < 48) y = c == 45, c = getchar();
    while (c > 47) x = x * 10 + c - 48, c = getchar();
    return y ? -x : x;
  }
} buf;
const int N = 2e5 + 5;
set<int> t[N];
int d[N], f[N];
int main() {
  int n = buf, m = buf, k = buf;
  vector<array<int, 2>> e;
  set<array<int, 2>> c;
  for (int i = 0; i < m; ++i) {
    int u = buf, v = buf;
    if (u > v) swap(u, v);
    c.insert({u, v});
    e.push_back({u, v});
    t[u].insert(v);
    t[v].insert(u);
    ++d[u];
    ++d[v];
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (d[i] < k) q.push(i);
  int s = n;
  for (int i = m - 1; ~i; --i) {
    while (q.size()) {
      int u = q.front();
      q.pop();
      for (int v : t[u]) {
        c.erase({min(u, v), max(u, v)});
        t[v].erase(u);
        if (d[v]-- == k) q.push(v);
      }
      --s;
    }
    f[i] = s;
    if (c.count(e[i])) {
      int u = e[i][0], v = e[i][1];
      t[u].erase(v);
      t[v].erase(u);
      if (d[u]-- == k) q.push(u);
      if (d[v]-- == k) q.push(v);
    }
  }
  for (int i = 0; i < m; ++i) printf("%d\n", f[i]);
}

### 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>
using namespace std;
template <class T1, class T2>
inline void upd1(T1& a, const T2& b) {
  a = a < b ? a : b;
}
template <class T1, class T2>
inline void upd2(T1& a, const T2& b) {
  a = a > b ? a : b;
}
template <class T>
inline bool equ(const T& a, const T& b) {
  return !memcmp(&a, &b, sizeof a);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
struct ano {
  operator long long() {
    long long x = 0, y = 0, c = getchar();
    while (c < 48) y = c == 45, c = getchar();
    while (c > 47) x = x * 10 + c - 48, c = getchar();
    return y ? -x : x;
  }
} buf;
const int N = 2e5 + 5;
set<int> t[N];
int d[N], f[N];
int main() {
  int n = buf, m = buf, k = buf;
  vector<array<int, 2>> e;
  set<array<int, 2>> c;
  for (int i = 0; i < m; ++i) {
    int u = buf, v = buf;
    if (u > v) swap(u, v);
    c.insert({u, v});
    e.push_back({u, v});
    t[u].insert(v);
    t[v].insert(u);
    ++d[u];
    ++d[v];
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (d[i] < k) q.push(i);
  int s = n;
  for (int i = m - 1; ~i; --i) {
    while (q.size()) {
      int u = q.front();
      q.pop();
      for (int v : t[u]) {
        c.erase({min(u, v), max(u, v)});
        t[v].erase(u);
        if (d[v]-- == k) q.push(v);
      }
      --s;
    }
    f[i] = s;
    if (c.count(e[i])) {
      int u = e[i][0], v = e[i][1];
      t[u].erase(v);
      t[v].erase(u);
      if (d[u]-- == k) q.push(u);
      if (d[v]-- == k) q.push(v);
    }
  }
  for (int i = 0; i < m; ++i) printf("%d\n", f[i]);
}

```