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 = 205;
char s[N];
int dp[N][N];
int main() {
  int n, lineNum, i, j, k, a, b;
  scanf("%d%d%d", &lineNum, &a, &b);
  scanf("%s", s);
  n = strlen(s);
  fill((int*)dp, (int*)dp + N * N, -1);
  dp[0][0] = 0;
  for (i = a; i <= n; i++) {
    for (j = 1; j <= lineNum; j++) {
      for (k = a; k <= b && k <= i; k++) {
        if (dp[i - k][j - 1] != -1) {
          dp[i][j] = k;
          break;
        }
      }
    }
  }
  if (dp[n][lineNum] != -1) {
    string tmp;
    int curIndex = 0;
    for (i = lineNum; i > 0; i--) {
      tmp.assign(s + curIndex, dp[n - curIndex][i]);
      printf("%s\n", tmp.c_str());
      curIndex += dp[n - curIndex][i];
    }
  } else {
    printf("No solution\n");
  }
  printf("\n");
  return 0;
}
