#include "bits/stdc++.h" #define int long long using namespace std; using ll = long long; //using P = pair>; const ll INF = (1LL << 61); ll mod = 1000000007; struct mint { ll x; // typedef long long ll; mint(ll x = 0) :x((x%mod + mod) % mod) {} mint operator-() const { return mint(-x); } mint& operator+=(const mint a) { if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint a) { if ((x += mod - a.x) >= mod) x -= mod; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; } mint operator+(const mint a) const { mint res(*this); return res += a; } mint operator-(const mint a) const { mint res(*this); return res -= a; } mint operator*(const mint a) const { mint res(*this); return res *= a; } mint pow(ll t) const { if (!t) return 1; mint a = pow(t >> 1); a *= a; if (t & 1) a *= *this; return a; } // for prime mod mint inv() const { return pow(mod - 2); } mint& operator/=(const mint a) { return (*this) *= a.inv(); } mint operator/(const mint a) const { mint res(*this); return res /= a; } }; istream& operator>>(istream& is, mint& a) { return is >> a.x; } ostream& operator<<(ostream& os, const mint& a) { return os << a.x; } vectora = { 6,2,5,5,4,5,6,4,7,6 }; int f(int n, string &x) { if (x[0] == '-') { x.erase(x.begin()); int ans = f(n - 1, x); x = '-' + x; return 1 + ans; } if (x.size() == 1) { return a[x[0] - '0']; } else { char t = x.back(); string a2 = ""; a2 += t; x.pop_back(); int ans = f(n - 1, x); ans += f(n - 1, a2); x += t; return ans; } } int f2(int n, int x) { if (x < 0) { return 1 + f2(n - 1, -x); } else if (x <= 9) { return a[x]; } else { return f2(n - 1, x / 10) + f2(n - 1, x % 10); } } signed main() { ios::sync_with_stdio(false); cin.tie(0); string P, N; cin >> P >> N; int s = P.size(); if (N.size() > 4) { int now = f(s, P); for (int i = 1; i < 10000; i++) { now = f2(s, now); }cout << now << endl; } else { if (N.size() == 1 && N[0] == '0') { cout << P << endl; return 0; } int now = f(s, P); for (int i = 1; i <= atoi(N.c_str()) - 1; i++) { now = f2(s, now); }cout << now << endl; } return 0; }