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 dp[210][210];
char s[210];
int main() {
  int a, b, len, i, j, k;
  scanf("%d%d%d", &k, &a, &b);
  scanf("%s", s);
  len = strlen(s);
  for (i = a; i <= b; i++)
    if (len - i + 1 >= 0) dp[len - i + 1][1] = 1;
  for (i = len; i >= 1; i--)
    for (j = 1; j < k; j++)
      if (dp[i][j])
        for (int ii = a; ii <= b; ii++)
          if (i - ii >= 1) dp[i - ii][j + 1] = 1;
  if (dp[1][k]) {
    int now = 1;
    dp[len + 1][0] = 1;
    while (k--) {
      for (i = a; i <= b; i++)
        if (i + now <= len + 1 && dp[i + now][k]) {
          for (j = now; j < now + i; j++) printf("%c", s[j - 1]);
          puts("");
          now += i;
          break;
        }
    }
    return 0;
  }
  puts("No solution");
}
