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;
const int MAXN = 222;
int f[MAXN][MAXN];
char s[MAXN];
int k, a, b, n;
int dfs(int n, int k) {
  if (n == 0) return 0;
  dfs(n - f[n][k], k - 1);
  for (int i = n - f[n][k]; i < n; i++) printf("%c", s[i]);
  printf("\n");
  return 0;
}
int main() {
  scanf("%d %d %d\n", &k, &a, &b);
  gets(s);
  n = strlen(s);
  f[0][0] = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = a; j <= b; j++) {
      if (i - j < 0) break;
      int x = i - j;
      for (int z = 0; z < k; z++)
        if (f[x][z]) f[i][z + 1] = i - x;
    }
  }
  if (f[n][k] == 0) {
    puts("No solution");
    return 0;
  }
  dfs(n, k);
  return 0;
}
