/*
 *   Super slow (exponential) solution.  Just for testing.
 */
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
using namespace std ;
using ll = long long ;
using ull = unsigned long long ;
ll X, n, m ;
vector<pair<ll, int>> both ;
vector<int> pairs ;
ll dist(ll a, ll b) {
   return min((a-b+X)%X, (b-a+X)%X) ;
}
const ll MOD = 998244353 ;
const int CNT = 1<<20 ;
int main() {
   cin >> X >> n >> m ;
   both.resize(n+m) ;
   pairs.resize(n+m) ;
   char tmp;
   for (auto &v: both) {
      cin >> v.first >> tmp;
      v.second = tmp == 'P';
   }
   sort(both.begin(), both.end()) ;
   for (int dir=-1; dir<=1; dir += 2) {
      vector<int> hstack ;
      for (int need=n, at=0; need > 0; at=(at+dir+n+m)%(n+m)) {
         if (both[at].second == 1) {
            hstack.push_back(at) ;
         } else if (hstack.size()) {
            int p = hstack[hstack.size()-1] ;
            hstack.pop_back() ;
            if (pairs[p] / CNT * 2 == dir + 1) {
               pairs[p] += CNT + at ;
               pairs[at] += CNT + p ;
               need-- ;
            }
         }
      }
   }
   ll r = 1, rs = 0 ;
   for (int i=0; i<n+m; i++)
      if (pairs[i] / CNT == 1) {
         ll firstv=0, secondv=0 ;
         for (int h=i, p = pairs[h]-CNT; ; p=pairs[h]-2*CNT-p) {
            int h2 = pairs[p]-2*CNT-h ;
            secondv += dist(both[h2].first, both[p].first) ;
            if (pairs[h2] / CNT == 1)
               break ;
            h = h2 ;
         }
         int bcnt = 1 ;
         ll bestv = secondv ;
         for (int h=i, p = pairs[h]-CNT; ; p=pairs[h]-2*CNT-p) {
            int h2 = pairs[p]-2*CNT-h ;
            secondv -= dist(both[h2].first, both[p].first) ;
            firstv += dist(both[h].first, both[p].first) ;
            if (secondv + firstv < bestv) {
               bcnt = 1 ;
               bestv = secondv + firstv ;
            } else if (secondv + firstv == bestv)
               bcnt++ ;
            if (pairs[h2] / CNT == 1) {
               pairs[h2] = 0 ;
               break ;
            }
            h = h2 ;
         }
         r = (r * bcnt) % MOD ;
         rs += bestv ;
      }
   cout << rs << endl << r << endl ;
}