Problem:
Score : 300 points

Problem Statement

Takahashi participated in a contest on AtCoder.

The contest had N problems.

Takahashi made M submissions during the contest.

The i-th submission was made for the p_i-th problem and received the verdict S_i (AC or WA).

The number of Takahashi's correct answers is the number of problems on which he received an AC once or more.

The number of Takahashi's penalties is the sum of the following count for the problems on which he received an AC once or more: the number of WAs received before receiving an AC for the first time on that problem.

Find the numbers of Takahashi's correct answers and penalties.

Constraints

N, M, and p_i are integers.

1 \leq N \leq 10^5

0 \leq M \leq 10^5

1 \leq p_i \leq N

S_i is AC or WA.

Input

Input is given from Standard Input in the following format:

N M
p_1 S_1
:
p_M S_M

Output

Print the number of Takahashi's correct answers and the number of Takahashi's penalties.

Sample Input 1

2 5
1 WA
1 AC
2 WA
2 AC
2 WA

Sample Output 1

2 2

In his second submission, he received an AC on the first problem for the first time. Before this, he received one WA on this problem.

In his fourth submission, he received an AC on the second problem for the first time. Before this, he received one WA on this problem.

Thus, he has two correct answers and two penalties.

Sample Input 2

100000 3
7777 AC
7777 AC
7777 AC

Sample Output 2

1 0

Note that it is pointless to get an AC more than once on the same problem.

Sample Input 3

6 0

Sample Output 3

0 0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <bits/stdc++.h>
 
#if DEBUG
    #include "debug.h"
#else
    #define debug(args...)
#endif    

using namespace std;

typedef vector<int> vi;
typedef pair<int,int> ii;
typedef long long ll;


#define M 1000000 
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define fill(a,x) memset(a,x,sizeof(a))
#define F first
#define S second
#define FOR(i,a,b) for(int i = a; i<=b; ++i)
#define NFOR(i,a,b) for(int i = a; i>=b; --i)
#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = INT_MAX;
const int mod = 1e9+7;

const int N = 2e5 + 10;
inline int add(int x,int y){
  x += y;
  if(x >= mod) x -= mod;
  return x;
}
inline int mul(int x,int y){
  x = (1LL * x * y) % mod;
  return x;
}
clock_t time_p=clock();

vector<string> sub[N];

int32_t main(){
  fast;

  //freopen("in.txt","r",stdin);
  //freopen("out.txt","w",stdout);

  int n,m;
  cin >> n >> m;
  FOR(i,1,m) {
    int p;
    string s;
    cin >> p >> s;
    sub[p].pb(s);
  }
  int a1 = 0, a2 = 0;
  FOR(i,1,n) {
    for(int j = 0; j < sz(sub[i]); j++) {
      if(sub[i][j] == "AC") {
        a1++;
        a2+= j;
        break;
      }
    }
  }
  cout << a1 << " " << a2 << "\n";

  time_p=clock()-time_p;
  cerr<<"\nTime Taken : "<<(float)(time_p)/CLOCKS_PER_SEC<<"\n";

  return 0;
}