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: Wrong Answer)

//  Created by conan1024hao in 2019.
//  Copyright © 2019 conan1024hao. All rights reserved.
//  専用ライブラリです、自由にコピーして構いません。
//  感谢看我的代码！Wechat:conan1024hao Twitter/QQ:810396815
#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 10e8
#define INF 10e16
#define MOD 1000000007
#define mod 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
//using Int=int_fast64_t;
typedef long long int ll;
typedef pair<ll,ll>Pll;
typedef pair<int,int>Pin;
long long GCD(long long a, long long b) { return b ? GCD(b, a%b) : a; }
long long LCM(long long a, long long b)  {return a/GCD(a,b)*b;}
int dx[8]={-1,0,1,0,1,1,-1,-1};
int dy[8]={0,-1,0,1,1,-1,1,-1};
char dir[4]={'u','l','d','r'};
ll cmp1(pair<ll,string>a,pair<ll,string> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se<b.se;
}

//---------------------------------------------------------------------------
long long modpow(long long a, long long n) {//a^n(MOD)を求める
    long long res = 1;
    while (n > 0) {
         if (n & 1) res = res * a % MOD;
         a= a * a % MOD;//次の準備
         n >>= 1;
      }
 return res;
}
//---------------------------------------------------------------------------
int main(){//問題をちゃんと見ろ!!!!!!!!!!!!!!!!!      llか??????????        memset()!!!!!!!!!!!!       ペナを減らせ!!!!!!!!!!!!!
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------
    int n,m;cin>>n>>m;
    int ac[100005];
    memset(ac,0,sizeof(ac));
    int ans=0,wa=0;
    for(int i=0;i<m;i++){
        int p;cin>>p;p--;
        //if(ac[p])continue;
        string s;cin>>s;
        if(s=="AC"){
            if(ac[p]==0){
                ac[p]++;
                ans++;
            }
        }
        else{
            if(ac[p]==0)wa++;
        }
        //cout<<ans<<" "<<wa<<Endl;
    }
    
    cout<<ans<<" "<<wa<<Endl;
    //-------------------------------
    return 0;
}
//---------------------------------------------------------------------------