text: # 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.# Solution #include <bits/stdc++.h> using namespace std; const int module = 1000000007; int main() { ios_base::sync_with_stdio(false); cin.tie(0); cout.