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 <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <cstdlib>
#include <map>
#include <fstream>
#include <string>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <random>
#include <set>
#include <queue>
#include <bitset>


using namespace std;

long long dp[3][100100] = {};
const long long INF = 1LL << 60;
#define MOD 1000000007
#define Int int64_t
#define PI 3.14159265358979
#define dump(a) cout << a << endl;

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

// mod. m での a の逆元 a^{-1} を計算する

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

/**
* @fn
* 最大値を取ります.Vector<int>を渡すこと.
*/
long long int imax(vector<int> a) {
    int MAX = -10000000;
    for (int i = 0; i < a.size(); i++) {
        if (MAX < a.at(i)) MAX = a.at(i);
    }
    return MAX;
}

/**
* @fn
* 最小値を取ります.Vector<int>を渡すこと.
*/
long long int imin(vector<int> a) {
    int MIN = 1000000000;
    for (int i = 0; i < a.size(); i++) {
        if (MIN > a.at(i)) MIN = a.at(i);
    }
    return MIN;
}

/**
* @fn
* 自然数について、桁数を求める
*/
Int getDigit(Int num) {
    return (Int) (log10(num)) + 1;
}


int get2Digit(Int num) {
    int cnt = 0;
    while (num > 0) {
        if (num % 2 != 0) {
            return cnt;
        }
        cnt++;
        num /= 2;
    }
    return cnt;
}

long long int absmin(vector<int> a, long long int b) {
    int mini = 100000000;
    int d = 0;
    for (int i = 0; i < a.size(); i++) {
        if (mini > abs(a.at(i) - b)) {
            mini = abs(a.at(i) - b);
            d = i;
        }
    }
    return d;
}


long long int absmax(vector<int> a, long long int b) {
    int max = -100000000;
    int d = 0;
    for (int i = 0; i < a.size(); i++) {
        if (max < abs(a.at(i) - b)) {
            max = abs(a.at(i) - b);
            d = i;
        }
    }
    return d;
}

Int gcd(Int a, Int b) {
    if (a < b) swap(a, b);
    if (b < 1) return -1;
    if (a % b == 0) return b;
    return gcd(b, a % b);
}

void haerobatya(int start, int end, char level) {
    if (start > end) swap(start, end);
    for (int i = start; i <= end; i++) {
        if (i < 100) {
            cout << "https://atcoder.jp/contests/abc0" << i << "/tasks/abc0" << i << "_" << level << endl;
        } else {
            cout << "https://atcoder.jp/contests/abc" << i << "/tasks/abc" << i << "_" << level << endl;
        }
    }
}

long long int cntmod(long long int warareru, long long int mod) {
    long long int i = 0;
    for (i = 0; true; i++) {
        if (warareru % mod == 0 && warareru >= mod) {
            warareru /= mod;
        } else {
            return i;
        }
    }
}

void ansYes(bool a) {
    if (a) cout << "Yes" << endl;
    else cout << "No" << endl;
}

void ansYES(bool a) {
    if (a) cout << "YES" << endl;
    else cout << "NO" << endl;
}


Int euclidean_gcd(Int a, Int b) {
    while (true) {
        if (a < b) swap(a, b);
        if (!b) break;
    }
    return a;
}

//UnionFind群
class UnionFind {
protected:
    vector<int> par;
    vector<int> data;
    int cnt;
public:
    UnionFind() : par(1), data(1), cnt(1) {};

    UnionFind(int N);

    UnionFind(const vector<int> &par, const vector<int> &data, int cnt);

    int root(int x);

    int marge(int x, int y);

    bool same(int x, int y);

    int add();

    int add(int x);

    inline int getSize(int x) { return data.at(root(x)); };

    int allroot();

    inline bool isroot(int x) { return (par.at(x) == -1); };
};


UnionFind::UnionFind(int N) : par(N), data(N), cnt(N) {
    for (int i = 0; i < N; i++) {
        par.at(i) = -1;
        data.at(i) = 1;
    }
}


int UnionFind::root(int x) {
    if (par.at(x) == -1) return x;
    return par.at(x) = root(par.at(x)); // 経路圧縮
}


int UnionFind::marge(int x, int y) {
    int rx = root(x);
    int ry = root(y);
    if (rx == ry) return rx;
    else { // ノード数の少ない方に接続
        if (data.at(rx) < data.at(ry)) swap(rx, ry);
        par.at(ry) = rx;
        data.at(rx) += data.at(ry);
        return rx;
    }
}

bool UnionFind::same(int x, int y) {
    return root(x) == root(y);
}

int UnionFind::add() {
    par.push_back(cnt);
    cnt++;
    return cnt - 1;
}

int UnionFind::add(int x) {
    par.push_back(cnt);
    cnt++;
    return marge(x, cnt);
}

int UnionFind::allroot() {
    int cnt = 0;
    for (const auto it : par) {
        if (it == -1) cnt++;
    }
    return cnt;
}


//約数の数を数える
int yakusu(int n) {
    int cnt = 0;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            cnt++;
            if (i * i != n) cnt++;
        }
    }
    return cnt;
}

int vecsum(vector<int> a) {
    int res = 0;
    for (auto z : a){
        res += z;
    }
    return res;
}

vector<int> yakusurekkyo(int n) {
    vector<int> a;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            a.push_back(i);
            if (i * i != n) a.push_back(n / i);
        }
    }
    return a;
}


bool IsPrime(int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}


vector<long long int> Prime;
void Eratosthenes(long long int n) {
    vector<bool> is_prime(n, true);
    for (int i = 2; i * i <= n; i++) { ;
        if (is_prime.at(i)) {
            for (int j = i + i; j < n; j += i) {
                is_prime.at(j) = false;
            }
        }
    }


    for (int i = 2;i < n;i++) {
        if (is_prime.at(i)){
            Prime.emplace_back(i);
        }
    }
}

int yokovec[4] = {0,-1,0,1};
int tatevec[4] = {1,0,-1,0};
int mp[500][500];
int h,w;
bool flag = false;

void bfs(int x,int y) {
    mp[x][y] = '#';
    for (int i = 0; i < 4;i ++){
        int tox = x + yokovec[i];
        int toy = y + tatevec[i];
        if (0 <= tox && tox < w && toy >= 0 && toy < w && mp[tox][toy] != '#') {
            if (mp[tox][toy] == 's') flag = true;
            bfs(tox,toy);
        }
    }
}


int main() {
    string s;
    cin >> s;
    string s2;
    cin >> s2;
    int cnt = 0;
    for (int i = 0; i < 3; i++) {
        if (s.at(i) == s2.at(i)) {
            cnt ++;
        }
    }
    cout << cnt << endl;
}