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 N = 500;
bool was[N][N];
int k, a, b;
int r[N];
char c[N];
bool ok(int n, int l) {
  if (n == 0 && l == 0) {
    int t = 0;
    for (int i = (int)k; i > 0; --i) {
      for (int j = 1; j <= (int)r[i]; ++j) printf("%c", c[t++]);
      puts("");
    }
    return true;
  }
  if (n == 0 || l == 0) return false;
  if (was[n][l]) return false;
  was[n][l] = true;
  for (int i = 1; i <= (int)l; ++i) {
    r[n] = i;
    if (i >= a && i <= b && ok(n - 1, l - i)) return true;
  }
  return false;
}
int main() {
  scanf("%d%d%d %s ", &k, &a, &b, c);
  if (!ok(k, strlen(c))) puts("No solution");
  return 0;
}
