#include<bits/stdc++.h>
#include<atcoder/all>
using namespace std;
using namespace atcoder;
typedef long long int ll;
typedef long double ld;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<vvvi> vvvvi;
typedef pair<ll,ll> pi;
typedef pair<ll,pi> ppi;
typedef pair<ll,ppi> pppi;
typedef pair<ll,pppi> ppppi;
#define FOR(i,l,r) for(ll i=l;i<r;i++)
#define REP(i,n) FOR(i,0,n)
#define RFOR(i,l,r) for(ll i=r-1;i>=l;i--)
#define RREP(i,n) RFOR(i,0,n)
#define ALL(x) x.begin(),x.end()
#define F first
#define S second
#define BS(A,x) binary_search(ALL(A),x)
#define LB(A,x) (ll)(lower_bound(ALL(A),x)-A.begin())
#define UB(A,x) (ll)(upper_bound(ALL(A),x)-A.begin())
#define COU(A,x) (UB(A,x)-LB(A,x))
#define sz(c) ((ll)(c).size())
template<typename T>using min_priority_queue=priority_queue<T,vector<T>,greater<T>>;
template<typename T1,typename T2>ostream&operator<<(ostream&os,pair<T1,T2>&p){os<<p.F<<" "<<p.S;return os;}
template<typename T1,typename T2>istream&operator>>(istream&is,pair<T1,T2>&p){is>>p.F>>p.S;return is;}
template<typename T>ostream&operator<<(ostream&os,vector<T>&v){REP(i,sz(v))os<<v[i]<<(i+1!=sz(v)?" ":"");return os;}
template<typename T>istream&operator>>(istream&is,vector<T>&v){for(T&in:v)is>>in;return is;}
template<class T>bool chmax(T&a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T&a,T b){if(b<a){a=b;return 1;}return 0;}
ld dist(ld x1,ld y1,ld x2,ld y2){return sqrtl((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}
//*
using mint=modint998244353;
const ll mod=998244353;
//*/
/*
using mint=modint1000000007;
const ll mod=1000000007;
//*/
//using mint=modint;
//*
typedef vector<mint> vm;
typedef vector<vm> vvm;
typedef vector<vvm> vvvm;
typedef vector<vvvm> vvvvm;
ostream&operator<<(ostream&os,mint&a){os<<a.val();return os;}
//*/
int main(){
  ll N,E;cin>>N>>E;
  if(E<11){
      ll a=1;
      REP(i,E)a*=5;
      REP(r,a)if((N-r*r)%a==0){cout<<r<<endl;return 0;}
      cout<<"NaN"<<endl;return 0;
  }
  ll a=1;
  REP(i,E)a*=5;
  N%=a;N+=a;N%=a;
  REP(x,1<<E){
    ll ans=0,s=1;
    REP(i,E){
        ll t=0;
        while((ans*ans-N)/s%5){t++;ans+=s;if(t>10){cout<<"NaN"<<endl;return 0;}}
        s*=5;
        if((x>>i)%2)ans=s-ans;
    }
    ans%=s;
    if(ans<=(1LL<<29)){cout<<ans<<endl;return 0;}
  }
  cout<<"NaN"<<endl;
  return 0;
}