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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000
#define MOD 1000000007
#define all(v) v.begin(),v.end()
#define pb push_back
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int kaijo[200010];
struct edge{int to,cost;};
int gcd(int a,int b){
	if(b==0)return a;
	return gcd(b,a%b);
}
int lcm(int a,int b){
	return a/gcd(a,b)*b;
}
bool prime(int a){
	if(a==1)return false;
	for(int i=2;i*i<=a;i++){
		if(a%i==0)return false;
	}
	return true;
}
bool kosa(double ax,double ay,double bx,double by,double cx,double cy,double dx,double dy){//線分の交差判定
	double ta=(cx-dx)*(ay-cy)+(cy-dy)*(cx-ax);
	double tb=(cx-dx)*(by-cy)+(cy-dy)*(cx-bx);
	double tc=(ax-bx)*(cy-ay)+(ay-by)*(ax-cx);
	double td=(ax-bx)*(dy-ay)+(ay-by)*(ax-dx);
	return tc*td<0&&ta*tb<0;
}
void init_fact(int n){
	kaijo[0]=1;
	for(int i=1;i<=n;i++){
		kaijo[i]=kaijo[i-1]*i;
		kaijo[i]%=MOD;
	}
}
int modpow(int a,int b){
	if(b==0)return 1;
	if(b%2)return modpow(a,b-1)*a%MOD;
	int memo=modpow(a,b/2);
	return memo*memo%MOD;
}
int comb(int a,int b){
	if(b>a)return 0;
	if(a==b)return 1;
	return kaijo[a]*modpow(kaijo[a-b],MOD-2)%MOD*modpow(kaijo[b],MOD-2)%MOD;
}
int g(int x){
	x=modpow(x,MOD-2);
	return x;
}
string s,t;
signed main(){
	cin>>s>>t;
	int ans=0;
	if(s[0]==t[0])ans++;
	if(s[1]==t[1])ans++;
	if(s[2]==t[2])ans++;
	cout<<ans<<endl;
	return 0;
}