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 long long N = 200 * 1000 + 9;
const long long inf = 1e18;
const long long mod = 1000000007;
void solve() {
  long long k, a, b;
  cin >> k >> a >> b;
  string s;
  cin >> s;
  long long n = (long long)s.size();
  long long cnt = 0;
  long long cn = 0;
  long long x;
  long long y = b - a;
  for (long long i = 0; i < n; i++) {
    cnt++;
    if (cnt == a) {
      cnt = 0;
      cn++;
      if (cn == k) {
        x = n - i - 1;
        if (a == b and i != n - 1) {
          cout << "No solution" << '\n';
          return;
        }
        if (a == b and i == n - 1) {
          long long cnt = 0;
          for (long long i = 0; i < n; i++) {
            cout << s[i];
            cnt++;
            if (cnt == a) {
              cnt = 0;
              cout << '\n';
            }
          }
          return;
        }
        break;
      }
    }
  }
  if (cn < k) {
    cout << "No solution" << '\n';
    return;
  }
  long long xx = x / y;
  long long yy = x % y;
  long long zz = xx;
  long long qq = 0;
  if (yy > 0) zz = xx + 1, qq = 1;
  if (zz > k) {
    cout << "No solution" << '\n';
    return;
  }
  yy = a + yy;
  long long pp = k - zz;
  cnt = 0;
  for (long long i = 0; i < n; i++) {
    cout << s[i];
    cnt++;
    if (pp) {
      if (cnt == a) {
        cout << '\n';
        pp--;
        cnt = 0;
      }
    } else if (qq) {
      if (cnt == yy) {
        cout << '\n';
        qq--;
        cnt = 0;
      }
    } else {
      if (cnt == b) {
        cnt = 0;
        cout << '\n';
      }
    }
  }
  return;
}
signed main() {
  solve();
  return 0;
}
