Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into k lines not shorter than a and not longer than b, for the text to resemble human speech more. Help Anfisa.

Input

The first line contains three integers k, a and b (1 ≤ k ≤ 200, 1 ≤ a ≤ b ≤ 200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols.

Output

Print k lines, each of which contains no less than a and no more than b symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes). 

Examples

Input

3 2 5
abrakadabra


Output

ab
rakad
abra


Input

4 1 2
abrakadabra


Output

No solution
#include <bits/stdc++.h>
using namespace std;
int N, k, a, b, dp[205][205], path[205][205];
char s[205];
void PrintPath(int len, int rem) {
  if (len == 0 && rem == 0) return;
  PrintPath(len - path[len][rem], rem - 1);
  for (int i = len - path[len][rem], cnt = 0; cnt < path[len][rem]; i++, cnt++)
    printf("%c", s[i]);
  printf("\n");
  return;
}
int main() {
  scanf("%d %d %d", &k, &a, &b);
  scanf("%s", &s);
  N = strlen(s);
  dp[0][0] = 1;
  for (int len = 1; len <= N; len++)
    for (int rem = 0; rem <= k; rem++) {
      if (a <= len && len <= b && rem == 1) {
        dp[len][rem] = 1;
        path[len][rem] = len;
        continue;
      }
      for (int cut = a; cut <= b && len >= cut; cut++)
        if (rem > 0 && dp[len - cut][rem - 1]) {
          dp[len][rem] = 1;
          path[len][rem] = cut;
          break;
        }
    }
  if (!dp[N][k])
    printf("No solution\n");
  else
    PrintPath(N, k);
  return 0;
}
