Problem: A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.

The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.

<image>

The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.

Your task is to divide the island between parties.

Input

The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.

Output

If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.

If there are several solutions output any.

Examples

Input

3 4 2 2 3
5 8 3


Output

YES
aaabb
aabbb
cbb..
ccb..


Input

3 2 1 4 4
1 2 3 4


Output

YES
abbd
cccd
...d
...d

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class P63D {

    static boolean multipleIndependent = false;

    void run() {
        int a = in.nextInt(), b = in.nextInt();
        int c = in.nextInt(), d = in.nextInt();
        int n = in.nextInt();
        int[] x = in.nextIntArray(n);
        int minH = Math.min(b, d);
        int maxH = Math.max(b, d);
        char[][] map = new char[maxH][a + c];
        for (char[] row : map) {
            Arrays.fill(row, '.');
        }
        int dir;
        if (b == d || (minH % 2 == 0 && b > d) || (minH % 2 == 1 && d > b)) {
            dir = 1;
        } else {
            dir = -1;
        }
        int curr = 0;
        for (int i = 0; i < minH; i++) {
            int j = dir > 0 ? 0 : a + c - 1;
            while ((dir > 0 && j < a + c) || (dir < 0 && j >= 0)) {
                map[i][j] = (char) ('a' + curr);
                if (--x[curr] == 0) {
                    curr++;
                }
                j += dir;
            }
            dir *= -1;
        }
        int l, r;
        if (dir > 0) {
            l = 0;
            r = a;
        } else {
            l = a;
            r = a + c;
        }
        for (int i = minH; i < maxH; i++) {
            int j = dir > 0 ? l : r - 1;
            while ((dir > 0 && j < r) || (dir < 0 && j >= l)) {
                map[i][j] = (char) ('a' + curr);
                if (--x[curr] == 0) {
                    curr++;
                }
                j += dir;
            }
            dir *= -1;
        }
        out.println("YES");
        for (char[] row : map) {
            out.println(String.valueOf(row));
        }
    }

    /* -----: Template :----- */
    static InputReader in;
    static PrintWriter out;

    public static void main(String[] args) {
        try (PrintWriter o = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)))) {
            in = new InputReader(System.in);
            out = o;
            int q = multipleIndependent ? in.nextInt() : 1;
            while (q-- > 0) {
                new P63D().run();
            }
        }
    }

    static Integer[] toWrapper(int[] arr) {
        Integer[] res = new Integer[arr.length];
        for (int i = 0; i < arr.length; i++) {
            res[i] = arr[i];
        }
        return res;
    }

    static Integer[][] toWrapper(int[][] arr) {
        Integer[][] res = new Integer[arr.length][];
        for (int i = 0; i < arr.length; i++) {
            res[i] = toWrapper(arr[i]);
        }
        return res;
    }

    static Long[] toWrapper(long[] arr) {
        Long[] res = new Long[arr.length];
        for (int i = 0; i < arr.length; i++) {
            res[i] = arr[i];
        }
        return res;
    }

    static Long[][] toWrapper(long[][] arr) {
        Long[][] res = new Long[arr.length][];
        for (int i = 0; i < arr.length; i++) {
            res[i] = toWrapper(arr[i]);
        }
        return res;
    }

    static class InputReader {

        private final BufferedReader reader;
        private StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            int bufferSize = 32 * 1024;
            reader = new BufferedReader(new InputStreamReader(stream), bufferSize);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public String nextLine() {
            if (true) {
                throw new RuntimeException("nextLine: not tested");
            }
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            StringBuilder sb = new StringBuilder();
            return tokenizer.nextToken("\n");
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int[] nextIntArray(int n) {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

        public long[] nextLongArray(int n) {
            long[] arr = new long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }

        public int[][] nextIntMap(int n, int m) {
            int[][] map = new int[n][m];
            for (int i = 0; i < n; i++) {
                map[i] = nextIntArray(m);
            }
            return map;
        }

        public long[][] nextLongMap(int n, int m) {
            long[][] map = new long[n][m];
            for (int i = 0; i < n; i++) {
                map[i] = nextLongArray(m);
            }
            return map;
        }

        public char[][] nextCharMap(int n) {
            char[][] map = new char[n][];
            for (int i = 0; i < n; i++) {
                map[i] = next().toCharArray();
            }
            return map;
        }
    }
}