Problem:
Score: 200 points

Problem Statement

We have a string S consisting of lowercase English letters.

If the length of S is at most K, print S without change.

If the length of S exceeds K, extract the first K characters in S, append ... to the end of them, and print the result.

Constraints

K is an integer between 1 and 100 (inclusive).

S is a string consisting of lowercase English letters.

The length of S is between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

K
S

Output

Print a string as stated in Problem Statement.

Sample Input 1

7
nikoandsolstice

Sample Output 1

nikoand...

nikoandsolstice has a length of 15, which exceeds K=7.

We should extract the first 7 characters in this string, append ... to the end of them, and print the result nikoand....

Sample Input 2

40
ferelibenterhominesidquodvoluntcredunt

Sample Output 2

ferelibenterhominesidquodvoluntcredunt

The famous quote from Gaius Julius Caesar.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

/*** Typedef ***/
typedef long long ll;
//typedef int ll;
typedef unsigned long long ull;
typedef    long double   ld;
///typedef    complex<ld>   P;
//#define    X   real()
//#define    Y   imag()
//ld restant(P p)
//{
//    return abs(p);
//}
//polar(x,y) mag=l,arg=y
//typedef    unordered_set<ll>   usll;
//typedef    unordered_multiset<ll>   umsll;

/*** Loops ***/
#define ffit(i, dat)    for (__typeof(dat.begin()) i = dat.begin(); i != dat.end(); ++i)
#define ff1(i,a,b) for(ll i = a; i < b; i++)
#define ff2(i,a,b) for(ll i = b-1; i >= a; i--)
#define ff3(i,a,b,m) for(ll i = a; i <b; i+=m)
#define ipar(ar,n) ll ar[n]; ff1(i,0,n) cin>>ar[i];
#define opar(ar,n) ff1(i,0,n) cout<<ar[i]<<" ";
#define foRev(num) for(ll i = num - 1; i >= 0; i--)
#define forIn(arr, num) for(ll i = 0; i < num; i++) sc1(arr[i]);
#define forIn1(arr, num) for(ll i = 1; i <= num; i++) sc1(arr[i]);
#define vpnt(ans) for(ll i = 0; i < ans.size(); i++) cout << ans[i] << (i + 1 < ans.size() ? ' ' : '\n');
#define apnt(arr, num) for(ll i = 0; i < num; i++) cout << arr[i] << (i + 1 < num ? ' ' : '\n');

/*** Define Values ***/

#define PI acos(-1.0)
#define eps 1e-7
#define size1 15

/*** STLs ***/
#define rsz resize
#define ins insert
#define ft front
#define bk back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define pb push_back
#define ff first
#define ss second
#define mp make_pair
#define mem(name, value) memset(name, value, sizeof(name))
#define MN3(n1,n2,n3) min(n1,min(n2,n3))
#define MN4(n1,n2,n3,n4) min(n1,min(n2,min(n3,n4)))
#define MX3(n1,n2,n3) max(n1,max(n2,n3))
#define MX4(n1,n2,n3,n4) max(n1,max(n2,max(n3,n4)))
#define mxar(a,n) *max_element(a,a+n)
#define mnar(a,n) *min_element(a,a+n)
#define arr_sm(a,n) accumulate(a,a+n,0)
#define vc_sm(ve) accumulate(ve.begin(),ve.end(),0)
#define mx_vc(ve) *max_element(ve.begin(),ve.end())
#define mn_vc(ve) *min_element(ve.begin(),ve.end())
//#define endl " \n"

/*** STLs ***/
typedef vector <ll> vll;
#define vvll vector<vll>
typedef set <ll> sll;
typedef multiset <ll> msll;
//typedef multimap <ll> mpll;
typedef queue <ll> qll;
typedef map <ll, ll> mll;
typedef pair <ll, ll> pll;
typedef vector <pair <ll, ll> > vpll;
typedef priority_queue <pair <ll, ll> > pqpll;
#define ml unordered_map<ll,ll>
///typedef    unordered_multiset<int>   umsi;
///typedef    unordered_set<int>   usi;
#define    oncnt    __builtin_popcount
#define    zerobegin   __builtin_clz
#define    zeroend   __builtin_ctz
#define    parity   __builtin_parity

///vector<vector<int>> v(10, vector<int>(20,500)); 2d vector initialization. of 10 rows and 20 columns, with value 500.

/*** Sorts ***/
#define all(v) (v).begin(), (v).end()
#define rev(v) reverse(all(v))
#define srt(v) sort(all(v))
#define srtgrt(v) sort(all(v), greater<ll>())
#define prsrt greater<pair<int,int>>()
/*** Some Prints ***/
#define ip(n) ll n; cin>>n;
#define ip2(n,k) ll n; cin>>n; ll k; cin>>k;
#define kl    endl
#define no cout << "NO" << '\n'
#define yes cout << "YES" << '\n'
#define case cout << "Case " << t++ << ": ";
#define ps(x,y)         fixed<<setprecision(y)<<x
#define inf             1e18
#define dbg(x) cout << '>' << #x << ':' << x << endl;
#define w(x)             ll x; cin>>x; while(x--)
/*bool isPerfectSquare(long double x)
{
    long double sr = sqrt(x);
    return ((sr - floor(sr)) == 0);
}

bool boundary(ll i,ll j,ll n, ll m)
{
    return (i>=0 && i<n && j>=0 &&  j<m);   //matrix boundary checking
}

ll ceil(const ll &a, const ll &b)
{
    return (a - 1)/b + 1;
}


string alpha="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";

int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};

bool prime[10000005];
void SieveOfEratosthenes(int n)
{
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1] = 0;
    for (int p=2; p*p<=n; p++)
    {
        if (prime[p] == true)
            for (int i=p*p; i<=n; i += p)
                prime[i] = false;
    }
}
ll bigmod(ll n,ll pow)
{
    if(pow==0)
        return 1;
    if(pow==1)
        return n%mod;

    ll ans=bigmod(n,pow/2);
    ans=(ans*ans)%mod;

    if(pow%2==1)
    {
        ans=(ans*n)%mod;
    }
    return ans%mod;
}
ll fact[1000005];

ll nCr(ll n,ll r)
{

    ll ans=fact[n];
    ans=(ans*bigmod(fact[r],mod-2))%mod;
    ans=(ans*bigmod(fact[n-r],mod-2))%mod;
    return ans;
}*/
ll power(ll x, ll y)
{
    ll temp;
    if(y == 0)
        return 1;
    temp = power(x, y / 2);
    if (y % 2 == 0)
        return temp * temp;
    else
    {
        if(y > 0)
            return (x * temp * temp);
        else
            return ((temp * temp) / x);
    }
}
#define mod 10000007
//const ll mod=1e6+7;
ll pmod(ll a,ll b){ll res=1;a%=mod;assert(b>=0);for(; b; b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll expo(ll x, ll y){ ll res = 1; x = x % mod;  while (y > 0) { if (y & 1)res = (res*x) % mod; y = y>>1;x = (x*x) % mod;}return res;}
void matmul(ll a[][2], ll b[][2]) { ll tmp[2][2]; for(int i=0; i<2; i++) { for(int j=0; j<2; j++) { tmp[i][j] = 0; for(int k=0; k<2; k++) tmp[i][j] = (tmp[i][j] + a[i][k]*b[k][j]) % mod; }} for(int i=0; i<2; i++) for(int j=0; j<2; j++) a[i][j] = tmp[i][j]; }
ll fibo(int n)    // returns nth fibonacci number, also S(n) = f(n+2)-1
{ if(n == 0) return 0; else if(n == 1) return 1; ll T[2][2] = { {0, 1} , {1, 1} }; ll res[2][2] = { {1, 0}, {0,1} }; n--; while(n) { if(n & 1) matmul(res, T); n >>= 1; matmul(T, T); } return res[1][1]; }

#define go(i, a) for (auto i : a)
#define go1(it, v) for(auto it = v.begin(); it != v.end(); ++it)
#define mk(arr,n,type) type *arr=new type[n];
#define sz(obj) (int(obj.size()))
#define ipve(vc,n) vll vc(n); ff1(i,0,n){cin>>vc[i];}
#define ipvc(vc,n) vll vc; ff1(i,0,n){ip(x);vc.pb(x);}
bool sortbysec(const pair<pair<ll,ll>,ll> &a,const pair<pair<ll,ll>,ll> &b)
{
    return (a.second < b.second);
}
bool cmp(pll a,pll b)
{
     return a.ff - a.second > b.ff - b.ss;
}
bool p2(ll n)
{
   if(n==0)
   return false;

   return (ceil(log2(n)) == floor(log2(n)));
}

const ll N = 2e5+5;
float hex(float a)
{

    // Side cannot be negative
    if (a < 0)
        return -1;

    // Side of the hexagon
    float x = a/0.5176;
    return x;
}
int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
   ip(n); string s;cin>>s;
   if(n>=sz(s))
   {
       cout<<s;
   }
   else
   {
       string s1 = s.substr(0,n),s2;
        ff1(i,0,sz(s)-n)
        {
            s2+=".";
        }
       cout<<s1+s2;
   }
 return 0;
}