You are given the string s of length n and the numbers p, q. Split the string s to pieces of length p and q.

For example, the string "Hello" for p = 2, q = 3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".

Note it is allowed to split the string s to the strings only of length p or to the strings only of length q (see the second sample test).

Input

The first line contains three positive integers n, p, q (1 ≤ p, q ≤ n ≤ 100).

The second line contains the string s consists of lowercase and uppercase latin letters and digits.

Output

If it's impossible to split the string s to the strings of length p and q print the only number "-1".

Otherwise in the first line print integer k — the number of strings in partition of s.

Each of the next k lines should contain the strings in partition. Each string should be of the length p or q. The string should be in order of their appearing in string s — from left to right.

If there are several solutions print any of them.

Examples

Input

5 2 3
Hello


Output

2
He
llo


Input

10 9 5
Codeforces


Output

2
Codef
orces


Input

6 4 5
Privet


Output

-1


Input

8 1 1
abacabac


Output

8
a
b
a
c
a
b
a
c
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<vector<T>>;
constexpr ll mod = 1000000007;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 + 99);
const ld PI = acos((ld)-1);
const vector<ll> dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1};
template <typename T, typename U>
inline bool chmin(T& t, const U& u) {
  if (t > u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T, typename U>
inline bool chmax(T& t, const U& u) {
  if (t < u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
inline void yes() { cout << "Yes" << '\n'; }
inline void no() { cout << "No" << '\n'; }
template <typename T, typename Y>
inline T mpow(T a, Y n) {
  T res = 1;
  for (; n; n >>= 1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}
template <typename T>
V<T> prefix_sum(const V<T>& v) {
  int n = v.size();
  V<T> ret(n + 1);
  for (ll i = 0; i < (ll)(n); ++i) ret[i + 1] = ret[i] + v[i];
  return ret;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& vec) {
  for (auto&& x : vec) is >> x;
  return is;
}
template <typename T, typename Y>
ostream& operator<<(ostream& os, const pair<T, Y>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const V<T>& v) {
  os << "{";
  for (auto e : v) os << e << ",";
  return os << "}";
}
template <typename... Args>
void debug(Args&... args) {
  for (auto const& x : {args...}) {
    cerr << x << ' ';
  }
  cerr << '\n';
}
signed main() {
  cin.tie(0);
  cerr.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  ll n, a, b;
  cin >> n >> a >> b;
  string s;
  cin >> s;
  for (ll i = 0; i < (ll)(n + 1); ++i) {
    if (a * i > n) break;
    if ((n - a * i) % b) continue;
    V<string> ans(0);
    ll now = 0;
    while (now < n) {
      if ((ll)(ans).size() < i) {
        ans.emplace_back(s.substr(now, a));
        now += a;
      } else {
        ans.emplace_back(s.substr(now, b));
        now += b;
      }
    }
    cout << (ll)(ans).size() << '\n';
    for (ll j = 0; j < (ll)((ll)(ans).size()); ++j) {
      cout << ans[j] << '\n';
    }
    return 0;
  }
  cout << -1 << '\n';
}
