Problem:
Score : 200 points

Problem Statement

You are given a string S consisting of lowercase English letters.
We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom.

Constraints

1 \leq w \leq |S| \leq 1000

S consists of lowercase English letters.

w is an integer.

Input

Input is given from Standard Input in the following format:

S
w

Output

Print the desired string in one line.

Sample Input 1

abcdefgh
3

Sample Output 1

adg

When we write down abcdefgh, starting a new line after every three letters, we get the following:

abc

def

gh

Concatenating the letters at the beginnings of these lines, we obtain adg.

Sample Input 2

lllll
1

Sample Output 2

lllll

Sample Input 3

souuundhound
2

Sample Output 3

suudon

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pi = pair<int, int>;
using vi = vector<int>;
using vii = vector<pi>;
#define endl '\n'
#define fast_io ios_base::sync_with_stdio(0); cin.tie(NULL);
#define all(a) (a).begin(),(a).end()
#define sz(a) int((a).size())
#define clr(a,x) memset((a),(x)sz(a))
#define present(t, x) (t.find(x) != t.end())
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define rep(i, x, y) for(decltype(y) i = (x) - ((x) > (y)); i != (y) - ((x) > (y)); i += 1 - 2 * ((x) > (y)))
#define forall(it, a) for(decltype((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define what_is(x) cerr << #x << " is " << (x) << endl;
#define db(x) cerr << __LINE__ << ": " << #x << " = " << (x) << endl
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#define time__(d) \
	for(long blockTime=(long)NULL;(blockTime==(long)NULL?(blockTime=clock())!=(long)NULL:false); \
	debug("%s:%.4fs\n",(d),(double)(clock()-blockTime)/CLOCKS_PER_SEC))
const int MOD = (int) 1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ld PI = acos((ld) -1); // pi
const ld EPS = 1e-9; // epsilon
#define eb emplace_back
#define ef emplace_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int imin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int imax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}

//time__("program time")

void solve(){

}

int main(){

	fast_io;
	//solve();
	string s;
	int w;
	cin >> s >> w;

	int n = sz(s);
	string ans = "";
	for(int i = 0;i < n;i += w){

		ans += s[i];
	}

	cout << ans << "\n";


return 0;
}