// Arup Guha
// 4/5/2015
// Solution to 2015 April USACO Silver Problem: Bessie Goes Moo
// Originally written in contest, commented later.

import java.util.*;
import java.io.*;

public class bgm {

	final public static String KEY = "BESIGOM";

	final public static String[] WORDS = {"BESSIE", "GOES", "MOO"};

	public static long[][] freq;
	public static int[] lookup;

	public static void main(String[] args) throws Exception {

		// Make look up table.
		lookup = new int[26];
		Arrays.fill(lookup, -1);
		for (int i=0; i<KEY.length(); i++)
			lookup[KEY.charAt(i)-'A'] = i;

		Scanner stdin = new Scanner(new File("bgm.in"));
		int n = stdin.nextInt();

		// Store relevant mod 7 frequencies.
		freq = new long[7][7];
		for (int i=0; i<n; i++) {
			int letter = (int)(stdin.next().charAt(0)-'A');
			int num = (stdin.nextInt()%7+7)%7;
			freq[lookup[letter]][num]++;
		}

		int[] myMod = new int[7];
		long res = f(myMod, 0);

		// Solve and write out the result.
		FileWriter fout = new FileWriter(new File("bgm.out"));
		fout.write(res+"\n");
		fout.close();
	}

	public static long f(int[] mod, int k) {

		if (k == 7)
			return eval(mod);

		// Try each possibility for slot k from 0 to 6 and add up the results.
		long res = 0L;
		for (int i=0; i<7; i++) {
			mod[k] = i;
			res += f(mod, k+1);
		}
		return res;
	}

	//(B+E+S+S+I+E)(G+O+E+S)(M+O+O) "BESIGOM";
	// ([0]+2*[1]+2[2]+[3])([4]+[5]+[1]+[2])([6]+2*[5]])
	public static long eval(int[] mod) {

		boolean ok = false;
		for (int i=0; i<WORDS.length; i++) {
			int cur = 0;
			for (int j=0; j<WORDS[i].length(); j++)
				cur += mod[lookup[WORDS[i].charAt(j)-'A']];
			if (cur%7 == 0) {
				ok = true;
				break;
			}
		}

		if (!ok) return 0L;

		long res = 1;
		for (int i=0; i<7; i++)
			res *= freq[i][mod[i]];
		return res;
	}
}