Problem: You are given a non-empty line s and an integer k. The following operation is performed with this line exactly once:

  * A line is split into at most k non-empty substrings, i.e. string s is represented as a concatenation of a set of strings s = t1 + t2 + ... + tm, 1 ≤ m ≤ k. 
  * Some of strings ti are replaced by strings tir, that is, their record from right to left. 
  * The lines are concatenated back in the same order, we get string s' = t'1t'2... t'm, where t'i equals ti or tir. 



Your task is to determine the lexicographically smallest string that could be the result of applying the given operation to the string s.

Input

The first line of the input contains string s (1 ≤ |s| ≤ 5 000 000), consisting of lowercase English letters. The second line contains integer k (1 ≤ k ≤ |s|) — the maximum number of parts in the partition.

Output

In the single line print the lexicographically minimum string s' which can be obtained as a result of performing the described operation. 

Examples

Input

aba
2


Output

aab


Input

aaaabacaba
2


Output

aaaaabacab


Input

bababa
1


Output

ababab


Input

abacabadabacaba
4


Output

aababacabacabad

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize(3)
#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 5e6 + 10, M = 1e7 + 10;
const int B = 19491001;
int n, k;
string s;
int l[M], r[M], tot = 0;
unsigned long long fn[N], bk[N], pw[N];
string len1(string s) {
  string t = s;
  reverse(t.begin(), t.end());
  if (t < s) return t;
  return s;
}
void Lyndon(int n, int m) {
  l[0] = r[0] = -1;
  tot = 0;
  for (int i = 0, j, k; i < n;) {
    j = i, k = i + 1;
    for (; j < m && k < m && s[j % n] <= s[k % n];) {
      if (s[j % n] == s[k % n])
        j++;
      else
        j = i;
      k++;
    }
    for (; i <= j;) {
      l[++tot] = i, r[tot] = i + (k - j) - 1;
      i += k - j;
    }
  }
  return;
}
void initpre(string s) {
  pw[0] = 1;
  for (int i = 1; i <= n + 2; i++) pw[i] = pw[i - 1] * B;
  for (int i = 1; i <= n; i++) {
    int c = s[i];
    fn[i] = fn[i - 1] * B + c;
  }
  for (int i = n; i >= 1; i--) {
    int c = s[i];
    bk[i] = bk[i + 1] * B + c;
  }
  return;
}
unsigned long long up(int l, int r) {
  return bk[l] - bk[r + 1] * pw[r - l + 1];
}
unsigned long long down(int l, int r) {
  return fn[r] - fn[l - 1] * pw[r - l + 1];
}
unsigned long long Val(int p, int l) {
  if (l <= p) return up(p - l + 1, p);
  return up(1, p) * pw[l - p] + down(p + 1, l);
}
char val(int p, int l) {
  if (l <= p) return s[p - l + 1];
  return s[l];
}
unsigned long long vAl(int p, int l) {
  if (l <= p) return down(1, l);
  return down(1, p) * pw[l - p] + up(n - (l - p) + 1, n);
}
char vaL(int p, int l) {
  if (l <= p) return s[l];
  return s[n - (l - p) + 1];
}
int cmp(int x, int y) {
  int l = 1, r = n, be = 0;
  int bf = min(n, 4);
  if (Val(x, bf) != Val(y, bf)) r = bf;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (Val(x, mid) == Val(y, mid))
      be = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return val(x, be + 1) < val(y, be + 1);
}
void midfind(string &mn) {
  int pos = 1;
  for (int i = 2; i <= n; i++) {
    if (s[i] <= s[pos] && cmp(i, pos)) {
      pos = i;
    }
  }
  string t = "";
  for (int i = pos; i >= 1; i--) t += s[i];
  for (int i = pos + 1; i <= n; i++) t += s[i];
  if (t < mn) mn = t;
  return;
}
int cmpf(int x, int y) {
  int l = min(x, y), r = n, be = l;
  int bf = min(n, l + 4);
  if (vAl(x, bf) != vAl(y, bf)) r = bf;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (vAl(x, mid) == vAl(y, mid))
      be = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return vaL(x, be + 1) < vaL(y, be + 1);
}
void frfind(string &mn) {
  int pos = 1;
  for (int i = 2; i <= n; i++) {
    if (cmpf(i, pos)) {
      pos = i;
    }
  }
  string t = "";
  for (int i = 1; i <= pos; i++) t += s[i];
  for (int i = n; i > pos; i--) t += s[i];
  if (mn > t) mn = t;
  return;
}
void mincir(string &mn) {
  reverse(s.begin(), s.end());
  Lyndon(n, n * 2);
  int pos = -1;
  for (int i = 1; i <= tot; i++) {
    if (l[i] < n) {
      pos = l[i];
    }
  }
  string in = "";
  for (int i = pos; i <= pos + n - 1; i++) in += s[i % n];
  reverse(s.begin(), s.end());
  if (mn > in) mn = in;
  return;
}
string len2() {
  string mn = len1(s);
  n = (int)s.length();
  mincir(mn);
  s = '0' + s + '0';
  initpre(s);
  midfind(mn);
  frfind(mn);
  return mn;
}
void solve(string s, string &a, string &b, int k) {
  a.clear(), b.clear();
  int cnt = 0, i, fg = 0, n = s.length();
  int pl = -1, pr = -1;
  for (i = tot; i >= 1; i--) {
    if (l[i] == r[i]) {
      pl = pr = -1;
      if (!fg) {
        if (cnt >= k) break;
        cnt++, fg = 1;
      }
      continue;
    }
    if (pl != -1) {
      if (down(pl + 1, pr + 1) == down(l[i] + 1, r[i] + 1)) {
        reverse(s.begin() + l[i], s.begin() + r[i] + 1);
        continue;
      }
    }
    if (cnt >= k) break;
    cnt++;
    reverse(s.begin() + l[i], s.begin() + r[i] + 1);
    fg = 0;
    pl = l[i], pr = r[i];
  }
  for (int j = n - 1; j > r[i]; j--) a += s[j];
  for (int j = r[i]; j >= 0; j--) b += s[j];
  return;
}
string work() {
  string a, b, t = s;
  if (k == 1) {
    return len1(s);
  }
  n = s.length();
  t = '0' + s;
  initpre(t);
  solve(s, a, b, k);
  if (b.empty()) return a;
  solve(s, a, b, k - 2);
  s = b;
  a += len2();
  return a;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> s >> k;
  reverse(s.begin(), s.end());
  n = s.length();
  Lyndon(n, n);
  string ans = work();
  cout << ans << endl;
  return 0;
}