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>
int l[5000010], r[5000010];
int lyndon(char zf[5000010], int n) {
  int i = 0, j = 0, k = 1, ans = 0, m = 0;
  while (i < n) {
    if (zf[k] == zf[j])
      j += 1, k += 1;
    else if (zf[k] > zf[j])
      j = i, k += 1;
    else {
      int l = k - j;
      while (i <= j) {
        ::l[m] = i;
        r[m++] = i + l - 1;
        i += l;
      }
      j = i;
      k = i + 1;
    }
  }
  return m;
}
char zf[5000010], od[5000010];
bool same(int a, int b, int l, int r) {
  if (b - a != r - l) return false;
  for (int i = a, j = l; i <= b; i++, j++) {
    if (zf[i] != zf[j]) return false;
  }
  return true;
}
bool same(int i, int j) { return same(l[i], r[i], l[j], r[j]); }
void output(int l, int r, bool rev = false) {
  if (rev) {
    for (int i = r; i >= l; i--) printf("%c", zf[i]);
  } else {
    for (int i = l; i <= r; i++) printf("%c", zf[i]);
  }
}
void strrev(char zf[5000010]) {
  int n = strlen(zf);
  for (int i = 0, j = n - 1; i < j; i++, j--) {
    char t = zf[i];
    zf[i] = zf[j];
    zf[j] = t;
  }
}
char db[10000010];
int zxbsf(char zf[5000010], int n) {
  for (int i = 0; i < n; i++) db[i] = db[i + n] = zf[i];
  int i = 0, j = 1, k;
  while (i < n && j < n) {
    for (k = 0; k < n && db[i + k] == db[j + k]; k++)
      ;
    if (k == n) break;
    char a = db[i + k], b = db[j + k];
    if (a > b)
      i += k + 1;
    else
      j += k + 1;
    if (i == j) j += 1;
  }
  return i < j ? i : j;
}
char jg[5000010], tm[5000010];
int bl[5000010], ne[10000010];
int strcmp(int x, int y, int r) {
  int i, j;
  for (i = bl[x], j = bl[y]; i <= r && j <= r; i++, j++) {
    if (zf[i] != zf[j]) return zf[i] - zf[j];
  }
  for (i = bl[x] - 1; j <= r; i--, j++) {
    if (zf[i] != zf[j]) return zf[i] - zf[j];
  }
  return 0;
}
int getlcp(char zf[10000010], int x, int y) {
  int jg = 0;
  while (zf[x + jg] == zf[y + jg]) jg += 1;
  return jg;
}
void exkmp(char zf[10000010], int n) {
  ne[0] = n;
  ne[1] = getlcp(zf, 0, 1);
  int k = 1, p = k + ne[k] - 1;
  for (int i = 2; i < n; i++) {
    int L = ne[i - k];
    if (i + L <= p)
      ne[i] = L;
    else {
      int z = p - i + 1;
      if (z < 0) z = 0;
      while (zf[z] == zf[i + z]) z += 1;
      ne[i] = z;
      k = i;
      p = k + z - 1;
    }
  }
}
void split(int r, int n, int m) {
  int w = zxbsf(zf, r + 1);
  zf[r + 1] = 0;
  strcpy(jg, zf);
  for (int i = 0; i <= r; i++) tm[i] = zf[(w + i) % (r + 1)];
  if (strcmp(tm, jg) < 0) strcpy(jg, tm);
  int k = 0, la = 0;
  for (int i = 0; i <= m; i++) {
    if (i > 0 && !same(i, i - 1)) {
      bl[k++] = la;
      la = ::r[i - 1] + 1;
    }
  }
  bl[k++] = la;
  while (k > 0) {
    if (strcmp(k, k - 1, r) < 0) break;
    k -= 1;
  }
  for (int i = bl[k]; i <= r; i++) tm[i - bl[k]] = zf[i];
  for (int i = bl[k] - 1, j = r + 1 - bl[k]; i >= 0; i--, j++) tm[j] = zf[i];
  if (strcmp(tm, jg) < 0) strcpy(jg, tm);
  int z = n - 1 - r;
  for (int i = z; i < n; i++) zf[i - z] = od[i];
  n -= z;
  zf[n] = 0;
  strcpy(tm, zf);
  if (strcmp(tm, jg) < 0) strcpy(jg, tm);
  strcpy(db, zf);
  strrev(db);
  for (int i = 0; i < n; i++) db[i + n] = zf[i];
  exkmp(db, n + n);
  int wz = 0;
  for (int i = 1; i < n; i++) {
    int z = ne[n + wz + 1];
    if (z < i - wz) {
      if (zf[wz + z + 1] < db[z]) wz = i;
      continue;
    }
    z = ne[i - wz];
    if (i + 1 + z < n && db[z] < db[i - wz + z]) wz = i;
  }
  for (int i = 0; i <= wz; i++) tm[i] = zf[i];
  for (int i = n - 1, j = wz + 1; i > wz; i--, j++) tm[j] = zf[i];
  if (strcmp(tm, jg) < 0) strcpy(jg, tm);
  printf("%s", jg);
}
void solve(int n, int k) {
  int m = lyndon(zf, n) - 1;
  while (k > 2 && m >= 0) {
    int zm = m;
    while (zm > 0 && same(m, zm - 1)) zm -= 1;
    k -= 1;
    output(l[zm], r[m]);
    if (zm > 0 && r[m] == l[m] && r[zm - 1] == l[zm - 1]) k += 1;
    m = zm - 1;
  }
  if (m < 0) return;
  split(r[m], n, m);
}
int main() {
  int n, k;
  scanf("%s%d", od, &k);
  n = strlen(od);
  for (int i = 0; i < n; i++) zf[n - i - 1] = od[i];
  if (k == 1) {
    if (strcmp(zf, od) < 0)
      printf("%s", zf);
    else
      printf("%s", od);
    return 0;
  }
  solve(n, k);
  return 0;
}