#pragma GCC target( "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")

#include <bits/stdc++.h>

using namespace std;

#define speed ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define sz(x) (int)(x.size())
#define all(x) x.begin(), x.end()

typedef long long ll;
typedef unsigned long long ull;


const ll INF = LLONG_MAX;
const int inf = INT_MAX;
const int N = 3e5+5;
const int MOD = 1e9+7;
const double EPS = 1e-9;
const int p = 31;

vector <ll> p_pow(N), hash_s(N);
ll hash_t, pp = 1;

void solve(){
    string s, t;
    cin >> s >> t;

    int n = sz(s);

    p_pow[0] = 1;

    for(int i = 1; i < n; i++)
        p_pow[i] = p_pow[i-1] * p;

    for(int i = 0; i < n; i++){
        hash_s[i] = (s[i] - 'a' + 1) * p_pow[i];
        if(i)
            hash_s[i] += hash_s[i-1];
    }

    for(int i = 0; i < sz(t); i++){
        hash_t += (t[i]-'a' + 1) * pp;
        pp *= 31;
    }

    //lets imagine that s = "ababa" and t = "ba"
    //and s[1..2] = t;

    int l = 1, r = 2;

    cout << hash_t << ' ' << hash_s[r] - hash_s[l] * pow(p, r-l+1);



}

int main(){
    speed;

    int T = 1;

    while(T--)
        solve();
    return 0;
}