#include <bits/stdc++.h>
#include <atcoder/modint>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
using mint = atcoder::modint998244353;
int main() {
  cin.tie(0); cout.tie(0);
  ios::sync_with_stdio(false);
  int N, K;
  cin >> N >> K;
  vector<int> u(N), v(N);
  vector<vector<pair<int, int>>> G(N);
  for(int i = 0; i < N; i++) {
    cin >> u[i] >> v[i];
    --u[i]; --v[i];
    G[u[i]].push_back(make_pair(v[i], i));
    G[v[i]].push_back(make_pair(u[i], i));
  }
  vector<int> seen(N, -1);
  vector<bool> used(N, false);
  seen[0] = 0;
  queue<int> que;
  que.push(0);
  while(!que.empty()) {
    int x = que.front();
    que.pop();
    for(pair<int, int> p : G[x]) {
      int y = p.first;
      int id = p.second;
      if(seen[y] == -1) {
        seen[y] = seen[x] + 1;
        que.push(y);
        used[id] = true;
      }
    }
  }
  vector<vector<int>> H(N);
  for(int i = 0; i < N; i++) {
    if(used[i]) {
      H[u[i]].emplace_back(v[i]);
      H[v[i]].emplace_back(u[i]);
    }
  }
  if(K == 2) {
    for(int i = 0; i < N; i++) {
      if(!used[i]) {
        if(seen[u[i]] % 2 != seen[v[i]] % 2) {
          cout << 2 << '\n';
        } else {
          cout << 0 << '\n';
        }
      }
    }
  } else {
    for(int i = 0; i < N; i++) {
      if(!used[i]) {
        vector<int> dep(N, -1);
        dep[u[i]] = 1;
        vector<vector<mint>> dp(N, vector<mint>(2));
        dp[u[i]][0] = 1;
        dp[u[i]][1] = 0;
        que.push(u[i]);
        while(!que.empty()) {
          int x = que.front();
          que.pop();
          for(int y : H[x]) {
            if(dep[y] == -1) {
              que.push(y);
              dep[y] = dep[x] + 1;
              dp[y][0] += dp[x][1];
              dp[y][1] += dp[x][0] * (K - 1);
              dp[y][1] += dp[x][1] * (K - 2);
            }
          }
        }
        int d = dep[v[i]];
        mint ans = mint(K) * mint(K - 1).pow(N - 1);
        mint rem = mint(K) * mint(K - 1).pow(N - d) * dp[v[i]][0];
        ans -= rem;
        cout << ans.val() << '\n';
      }
    }
  }
  return 0;
}