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 = 205;
int k, a, b;
char s[maxn];
vector<char> vec[205];
int main() {
  cin >> k >> a >> b;
  scanf("%s", s);
  int n = strlen(s);
  int minlength = a * k, maxlength = b * k;
  if (n > maxlength || n < minlength) {
    printf("No solution\n");
    return 0;
  }
  int num = n / k;
  int left = n - num * k;
  int idx = 0;
  for (int i = 0; i < k - left; i++) {
    for (int j = 0; j < num; j++) {
      vec[i].push_back(s[idx++]);
    }
  }
  for (int i = k - left; i < k; i++) {
    for (int j = 0; j < num + 1; j++) {
      vec[i].push_back(s[idx++]);
    }
  }
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < vec[i].size(); j++) {
      printf("%c", vec[i][j]);
    }
    printf("\n");
  }
  return 0;
}
