#include <iostream>
using namespace std;

const int MAX_R = 30;

int main() {
    int R, C, n;
    cin >> R >> C >> n;
    int w[MAX_R] = {};             // w[r]＝第 r 列從左數起，前 w[r] 行已經被堵住（一開始全 0）
    int placed = 0, discarded = 0;
    for (int i = 0; i < n; i++) {
        char t;
        int y;
        cin >> t >> y;             // 子題組 1 保證 t 一定是 B：一列三格
        if (w[y] + 3 > C) {        // 這一列剩下的空間連三格都塞不下
            discarded++;
        } else {
            w[y] += 3;
            placed += 3;
        }
    }
    cout << R * C - placed << ' ' << discarded << '\n';
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

// 五種形狀打成表：每個方格相對於「最上列、最右行」的位移 (dr, dc)
vector<pair<int, int>> shape(char type) {
    if (type == 'A') return {{0, 0}, {1, 0}, {2, 0}, {3, 0}};
    if (type == 'B') return {{0, -2}, {0, -1}, {0, 0}};
    if (type == 'C') return {{0, -1}, {0, 0}, {1, -1}, {1, 0}};
    if (type == 'D') return {{0, 0}, {1, -2}, {1, -1}, {1, 0}};
    return {{0, 0}, {1, -1}, {1, 0}, {2, -1}, {2, 0}};        // E
}

int R, C;
vector<vector<bool>> occupied;

// 把形狀的最右行放在第 right 行、最上列放在第 top 列，放得下嗎？
bool canPlace(int top, int right, const vector<pair<int, int>>& cells) {
    for (auto [dr, dc] : cells) {
        int r = top + dr, c = right + dc;
        if (r < 0 || r >= R || c < 0 || c >= C) return false;   // 出界
        if (occupied[r][c]) return false;                       // 撞到別的貨物
    }
    return true;
}

int main() {
    int n;
    cin >> R >> C >> n;
    occupied.assign(R, vector<bool>(C, false));
    int used = 0, discarded = 0;
    while (n--) {
        char type;
        int top;
        cin >> type >> top;
        vector<pair<int, int>> cells = shape(type);
        int right = C - 1;                          // 先試「剛好完全進入倉庫」的最右位置
        if (!canPlace(top, right, cells)) { discarded++; continue; }
        while (canPlace(top, right - 1, cells)) right--;   // 還能往左就繼續推
        for (auto [dr, dc] : cells) occupied[top + dr][right + dc] = true;
        used += cells.size();
    }
    cout << R * C - used << ' ' << discarded << '\n';
    return 0;
}

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

// 五種形狀「由上到下每一列有幾格」：每一列都靠右對齊，所以只記寬度就夠
vector<vector<int>> tiles{
    {1, 1, 1, 1},   // A
    {3},            // B
    {2, 2},         // C
    {1, 3},         // D
    {1, 2, 2}       // E
};
int grid_num[5] = {4, 3, 4, 4, 5};   // 每種形狀共幾格
int R, C;

// 把形狀 tile 的最上列放在第 y 列往左推：放得下就更新 w 並回傳 true，放不下回傳 false
bool push(vector<int> &tile, int y, vector<int> &w) {
    int ma = 0;                       // 推完之後整件貨物共同的右邊界（佔到第 ma - 1 行）
    for (int i = 0; i < (int)tile.size(); i++) {
        if (tile[i] + w[y + i] > C) return false;   // 這一列連最右邊都塞不下
        ma = max(ma, tile[i] + w[y + i]);
    }
    for (int i = 0; i < (int)tile.size(); i++) {
        w[y + i] = ma;                // 整件貨物一起停：這幾列的牆都推到 ma
    }
    return true;
}

int main() {
    int n;
    cin >> R >> C >> n;
    vector<int> w(R);                 // w[r]＝第 r 列從左數起前 w[r] 行已經到不了（一開始全 0）
    int ans = R * C;
    int drop_cnt = 0;
    for (int i = 0; i < n; i++) {
        char t;
        int y;
        cin >> t >> y;
        if (push(tiles[t - 'A'], y, w)) {
            ans -= grid_num[t - 'A'];
        } else {
            drop_cnt++;
        }
    }
    cout << ans << ' ' << drop_cnt << '\n';
}