Problem:
Score : 100 points

Problem Statement

You will be given a string S of length 3 representing the weather forecast for three days in the past.

The i-th character (1 \leq i \leq 3) of S represents the forecast for the i-th day. S, C, and R stand for sunny, cloudy, and rainy, respectively.

You will also be given a string T of length 3 representing the actual weather on those three days.

The i-th character (1 \leq i \leq 3) of S represents the actual weather on the i-th day. S, C, and R stand for sunny, cloudy, and rainy, respectively.

Print the number of days for which the forecast was correct.

Constraints

S and T are strings of length 3 each.

S and T consist of S, C, and R.

Input

Input is given from Standard Input in the following format:

S
T

Output

Print the number of days for which the forecast was correct.

Sample Input 1

CSS
CSR

Sample Output 1

2

For the first day, it was forecast to be cloudy, and it was indeed cloudy.

For the second day, it was forecast to be sunny, and it was indeed sunny.

For the third day, it was forecast to be sunny, but it was rainy.

Thus, the forecast was correct for two days in this case.

Sample Input 2

SSR
SSR

Sample Output 2

3

Sample Input 3

RRR
SSS

Sample Output 3

0

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
using static System.Console;

namespace AtCoder_1
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = Cin();
            var t = Cin();
            int c = 0;
            for(int i = 0;i < 3; i++)
            {
                if(N[i] == t[i])
                {
                    c++;
                }
            }
            Cout(c);
            Cin();
        }

        static int GCD(int a, int b)
        {
            if (a < b) { swap(ref a, ref b); }
            if (a % b == 0) { return b; }
            else { return GCD(b, a % b); }
        }

        static void swap<T>(ref T a, ref T b)
        {
            T t = a;
            a = b;
            b = t;
        }

        static readonly int MOD10E9 = (1000000000 + 7);
        static bool isInt(double d) { return (int)d == d; }
        static string Cin() { return ReadLine(); }
        static string[] CinAr() { return ReadLine().Split(' '); }
        static int CinT() { return S2i(Cin()); }
        static int[] CinTAr() { return CinAr().Select((x, i) => { return S2i(x); }).ToArray(); }
        static long CinL() { return S2L(Cin()); }
        static long[] CinLAr() { return CinAr().Select((x, i) => { return S2L(x); }).ToArray(); }
        static double CinD() { return S2D(Cin()); }
        static double[] CinDAr() { return CinAr().Select((x, i) => { return S2D(x); }).ToArray(); }
        static int S2i(string s) { return int.Parse(s); }
        static long S2L(string s) { return long.Parse(s); }
        static double S2D(string s) { return double.Parse(s); }
        static void Cout(object s) { WriteLine(s); }
        static int[] yakusuu(int n)
        {
            if (n == 1) { return new int[] { 1 }; }
            HashSet<int> res = new HashSet<int>();
            for (int i = 1; i < Math.Ceiling(n / 2.0); i++)
            {
                if (n % i == 0)
                {
                    res.Add(i);
                    res.Add(n / i);
                }
            }
            return res.ToArray();
        }
        static void CoutAr<Type>(IEnumerable<Type> types)
        {
            string text = string.Join(" ", types);
            WriteLine(text);
        }

        public static long nCr(int n, int r)
        {
            // naive: return Factorial(n) / (Factorial(r) * Factorial(n - r));
            return nPr(n, r) / Factorial(r);
        }

        public static long nPr(int n, int r)
        {
            // naive: return Factorial(n) / Factorial(n - r);
            return FactorialDivision(n, n - r);
        }

        private static long FactorialDivision(int topFactorial, int divisorFactorial)
        {
            long result = 1;
            for (int i = topFactorial; i > divisorFactorial; i--)
                result *= i;
            return result;
        }

        private static long Factorial(int i)
        {
            if (i <= 1)
                return 1;
            return i * Factorial(i - 1);
        }
    }
}