There are n cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".

Find the number of cows that can currently show their hands without affecting any betting decisions.

Input

The first line contains a single integer, n (2 ≤ n ≤ 2·105). The second line contains n characters, each either "A", "I", or "F". The i-th character is "A" if the i-th player's status is "ALLIN", "I" if the i-th player's status is "IN", or "F" if the i-th player's status is "FOLDED".

Output

The first line should contain a single integer denoting the number of players that can currently show their hands.

Examples

Input

6
AFFAAA


Output

4


Input

3
AFI


Output

1

Note

In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
import java.io.*;
import java.util.*;


public class test {
    static boolean DEBUG_FLAG = false;
    int INF = (int)1e9;
    long MOD = 1000000007;
    
    static void debug(String s) {
        if(DEBUG_FLAG) {
            System.out.print(s);
        }
    }
    
    
    void solve(InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        String s = in.next();
        int a = 0, f = 0, i = 0;
        for(int j=0; j<n; j++) {
            if(s.charAt(j)=='A') a++;
            else if(s.charAt(j)=='F') f++;
            else i++;
        }
        int rs = 0;
        if(i==0) {
            rs = a;
        } else if(i==1) {
            rs = 1;
        }
        out.println(rs);
    }
    
    
    public static void main(String[] args) throws IOException {
        if(args.length>0 && args[0].equalsIgnoreCase("d")) {
            DEBUG_FLAG = true;
        }
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt();
        long start = System.nanoTime();
        while(t-- >0) {
            new test().solve(in, out);
        }
        long end = System.nanoTime();
        debug("\nTime: " + (end-start)/1e6 + " \n\n");
        out.close();
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
    
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}