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>
using namespace std;
const int _ = 5e6 + 7;
char str[_], tmp[_];
int L, K, pos;
vector<pair<int, int> > now;
vector<int> len;
void Lyndon() {
  int k = 1, i = 1, j = 1;
  while (k <= L)
    if (++j > L || tmp[k + (j - k) % (i - k + 1)] > tmp[j]) {
      int L = i - k + 1;
      len.push_back(L);
      int l = k;
      while (i < j) {
        i += L;
        k += L;
      }
      now.push_back(make_pair(l, k));
      i = j = k;
    } else if (tmp[k + (j - k) % (i - k + 1)] < tmp[j])
      i = j;
  for (pos = now.size() - 1; ~pos && K > 2; --K, --pos)
    if (len[pos] == 1)
      while (pos && len[pos - 1] == 1) --pos;
    else
      reverse(str + L - now[pos].second + 2, str + L - now[pos].first + 2);
  memcpy(tmp, str, sizeof(tmp));
  reverse(tmp + 1, tmp + L + 1);
}
char ans[_];
int t;
void solve1() {
  int l = L - t + 2, r = L;
  memcpy(ans, str, sizeof(ans));
  while (l <= r)
    if (str[l] == str[r]) {
      ++l;
      --r;
    } else {
      if (str[l] > str[r]) reverse(ans + L - t + 2, ans + L + 1);
      break;
    }
}
void solve00() {
  int i = 1, j = 2;
  while (max(i, j) < t) {
    int l = -1;
    while (++l < t)
      if (tmp[(i + l - 1) % (t - 1) + 1] != tmp[(j + l - 1) % (t - 1) + 1])
        break;
    if (l == t) break;
    if (tmp[(i + l - 1) % (t - 1) + 1] < tmp[(j + l - 1) % (t - 1) + 1])
      j += l + 1;
    else
      i += l + 1;
    if (i == j) ++j;
  }
  int l1 = L - t + 2, l2 = L - min(i, j) + 1;
  reverse(str + l1, str + l2 + 1);
  reverse(str + l2 + 1, str + L + 1);
  if (strcmp(str + 1, ans + 1) < 0) memcpy(ans, str, sizeof(ans));
  reverse(str + l1, str + l2 + 1);
  reverse(str + l2 + 1, str + L + 1);
}
int z[_], lcp[_];
void solve01() {
  int mx = 1, id = 1;
  for (int i = 2; i <= L; ++i) {
    z[i] = max(min(mx - i, z[i - id + 1]), 0);
    while (tmp[z[i] + 1] == tmp[z[i] + i]) ++z[i];
    if (i + z[i] > mx) {
      mx = i + z[i];
      id = i;
    }
  }
  mx = id = 1;
  for (int i = 1; i <= L; ++i) {
    lcp[i] = max(min(mx - i, z[i - id + 1]), 0);
    while (lcp[i] + i <= L && str[lcp[i] + i] == tmp[lcp[i] + 1]) ++lcp[i];
    if (i + lcp[i] > mx) {
      mx = i + lcp[i];
      id = i;
    }
  }
  int pos = L - t + 2;
  for (int i = pos + 1; i <= L; ++i)
    if (lcp[pos] < i - pos)
      pos = str[pos + lcp[pos]] < tmp[lcp[pos] + 1] ? i : pos;
    else {
      int t = z[i - pos + 1];
      pos = tmp[i - pos + 1 + t] < tmp[1 + t] ? pos : i;
    }
  reverse(str + pos, str + L + 1);
  if (strcmp(str + 1, ans + 1) < 0) memcpy(ans, str, sizeof(ans));
  reverse(str + pos, str + L + 1);
}
bool checkprefix(int l1, int l2) {
  if (2 * (t - l2) >= t - l1) return 0;
  for (int i = 0; i + l2 < t; ++i)
    if (tmp[i + l1] != tmp[i + l2]) return 0;
  return 1;
}
void solve10() {
  int p = pos, l1, l2;
  while (p && checkprefix(now[p - 1].first, now[p].first)) --p;
  int mn = now[p].first;
  for (int i = p + 1; i <= pos; ++i) {
    l1 = now[i - 1].first;
    l2 = now[i].first;
    int p1 = l1 + (t - l2), p2 = l2 - 1;
    bool flg = 0;
    while (p1 < t && !flg)
      if (tmp[p1] == tmp[p2]) {
        ++p1;
        --p2;
      } else if (tmp[p1] > tmp[p2])
        flg = 1;
      else
        break;
    if (flg) mn = now[i].first;
  }
  reverse(str + L - t + 2, str + L - mn + 2);
  if (strcmp(str + 1, ans + 1) < 0) memcpy(ans, str, sizeof(ans));
}
int main() {
  scanf("%s %d", str + 1, &K);
  L = strlen(str + 1);
  memcpy(tmp, str, sizeof(tmp));
  reverse(tmp + 1, tmp + L + 1);
  Lyndon();
  if (pos >= 0) {
    t = now[pos].second;
    solve1();
    if (K != 1 && pos) {
      solve00();
      solve01();
      solve10();
    }
  } else
    memcpy(ans, str, sizeof(ans));
  puts(ans + 1);
  return 0;
}