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;
template <class T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
T my_pow(T n, T p) {
  if (p == 0) return 1;
  T x = my_pow(n, p / 2);
  x = (x * x);
  if (p & 1) x = (x * n);
  return x;
}
template <class T>
T big_mod(T n, T p, T m) {
  if (p == 0) return (T)1;
  T x = big_mod(n, p / 2, m);
  x = (x * x) % m;
  if (p & 1) x = (x * n) % m;
  return x;
}
template <class T>
inline T Mod(T n, T m) {
  return (n % m + m) % m;
}
template <class T>
T extract(string s, T ret) {
  stringstream ss(s);
  ss >> ret;
  return ret;
}
template <class T>
string itos(T n) {
  ostringstream ost;
  ost << n;
  ost.flush();
  return ost.str();
}
long long stoi(string s) {
  long long r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
double toDouble(string s) {
  double r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
long long ar[2000005], vis[200005];
long long a = 0, b = 0, c = 0, r = 0, rr = 0, res = 0, n, m;
string s;
int main() {
  cin >> n >> a >> b;
  cin >> s;
  if (a + b == n) {
    cout << 2 << endl;
    cout << s.substr(0, a) << endl;
    cout << s.substr(a, b) << endl;
    return 0;
  }
  long long aa = s.size();
  if (aa % a == 0) {
    cout << aa / a << endl;
    c = 0;
    for (__typeof(aa / a) i = 0; i < (aa / a); i++) {
      cout << s.substr(c, a) << endl;
      c += (a);
    }
    return 0;
  }
  if (aa % b == 0) {
    cout << aa / b << endl;
    c = 0;
    for (__typeof(aa / b) i = 0; i < (aa / b); i++) {
      cout << s.substr(c, b) << endl;
      c += (b);
    }
    return 0;
  }
  for (__typeof(n) i = 1; i <= (n); i++) {
    long long cc = n - i * a;
    if (cc > 0 and cc % b == 0) {
      cout << i + cc / b << endl;
      c = 0;
      for (__typeof(i) j = 0; j < (i); j++) {
        cout << s.substr(c, a) << endl;
        c += (a);
      }
      for (__typeof(cc / b) i = 0; i < (cc / b); i++) {
        cout << s.substr(c, b) << endl;
        c += (b);
      }
      return 0;
    }
  }
  for (__typeof(n) i = 1; i <= (n); i++) {
    long long cc = n - i * b;
    if (cc > 0 and cc % a == 0) {
      cout << i + cc / a << endl;
      c = 0;
      for (__typeof(i) j = 0; j < (i); j++) {
        cout << s.substr(c, b) << endl;
        c += (b);
      }
      for (__typeof(cc / b) i = 0; i < (cc / b); i++) {
        cout << s.substr(c, a) << endl;
        c += (a);
      }
      return 0;
    }
  }
  cout << -1;
  return 0;
}
