You are wandering in the explorer space of the 2050 Conference.

The explorer space can be viewed as an undirected weighted grid graph with size n× m. The set of vertices is \{(i, j)|1≤ i≤ n, 1≤ j≤ m\}. Two vertices (i_1,j_1) and (i_2, j_2) are connected by an edge if and only if |i_1-i_2|+|j_1-j_2|=1.

At each step, you can walk to any vertex connected by an edge with your current vertex. On each edge, there are some number of exhibits. Since you already know all the exhibits, whenever you go through an edge containing x exhibits, your boredness increases by x.

For each starting vertex (i, j), please answer the following question: What is the minimum possible boredness if you walk from (i, j) and go back to it after exactly k steps?

You can use any edge for multiple times but the boredness on those edges are also counted for multiple times. At each step, you cannot stay on your current vertex. You also cannot change direction while going through an edge. Before going back to your starting vertex (i, j) after k steps, you can visit (i, j) (or not) freely.

Input

The first line contains three integers n, m and k (2≤ n, m≤ 500, 1≤ k≤ 20).

The j-th number (1≤ j ≤ m - 1) in the i-th line of the following n lines is the number of exibits on the edge between vertex (i, j) and vertex (i, j+1). 

The j-th number (1≤ j≤ m) in the i-th line of the following n-1 lines is the number of exibits on the edge between vertex (i, j) and vertex (i+1, j). 

The number of exhibits on each edge is an integer between 1 and 10^6.

Output

Output n lines with m numbers each. The j-th number in the i-th line, answer_{ij}, should be the minimum possible boredness if you walk from (i, j) and go back to it after exactly k steps.

If you cannot go back to vertex (i, j) after exactly k steps, answer_{ij} should be -1. 

Examples

Input


3 3 10
1 1
1 1
1 1
1 1 1
1 1 1


Output


10 10 10
10 10 10
10 10 10


Input


2 2 4
1
3
4 2


Output


4 4
10 6


Input


2 2 3
1
2
3 4


Output


-1 -1
-1 -1

Note

In the first example, the answer is always 10 no matter how you walk.

In the second example, answer_{21} = 10, the path is (2,1) → (1,1) → (1,2) → (2,2) → (2,1), the boredness is 4 + 1 + 2 + 3 = 10.
//package Codeforces.Round718Div1_2;

import java.io.*;
import java.util.*;

public class D {
    static boolean isValid(int n, int m, int i, int j){
        return 0<=i && i<n && 0<=j && j<m;
    }
    public static void main(String[] args) throws IOException {
        Soumit sc = new Soumit();

        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();

        StringBuilder sb = new StringBuilder();
        if(k%2==1){
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    sb.append("-1 ");
                }
                sb.append("\n");
            }
            System.out.println(sb);
            System.exit(0);
        }

        k/=2;

        long[][] horizontaledge = new long[n][m-1];
        long[][] verticaledge = new long[n-1][m];
        for(int i=0;i<n;i++)
            horizontaledge[i] = sc.nextLongArray(m-1);
        for(int i=0;i<n-1;i++)
            verticaledge[i] = sc.nextLongArray(m);

        long[][][] dp = new long[11][n][m];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                dp[0][i][j] = 0;
            }
        }
        for(int i=1;i<=k;i++){
            for(int j1=0;j1<n;j1++){
                for(int j2=0;j2<m;j2++){
                    long min = Long.MAX_VALUE/2000;
                    //for up
                    if(isValid(n, m, j1-1, j2)){
                        min = Math.min(dp[i-1][j1-1][j2]+verticaledge[j1-1][j2], min);
                    }

                    //for down
                    if(isValid(n, m, j1+1, j2)){
                        min = Math.min(min, dp[i-1][j1+1][j2]+verticaledge[j1][j2]);
                    }

                    //for left
                    if(isValid(n, m, j1, j2-1)){
                        min = Math.min(min, dp[i-1][j1][j2-1]+horizontaledge[j1][j2-1]);
                    }

                    //for right
                    if(isValid(n, m, j1, j2+1)){
                        min = Math.min(min, dp[i-1][j1][j2+1]+horizontaledge[j1][j2]);
                    }

                    dp[i][j1][j2] = min;
                }
            }
        }

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                sb.append(dp[k][i][j]*2).append(" ");
            }
            sb.append("\n");
        }

        System.out.println(sb);

        sc.close();
    }

    static class Soumit {
        final private int BUFFER_SIZE = 1 << 18;
        final private DataInputStream din;
        final private byte[] buffer;
        private PrintWriter pw;
        private int bufferPointer, bytesRead;
        StringTokenizer st;

        public Soumit() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Soumit(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public void streamOutput(String file) throws IOException {
            FileWriter fw = new FileWriter(file);
            BufferedWriter bw = new BufferedWriter(fw);
            pw = new PrintWriter(bw);
        }

        public void println(String a) {
            pw.println(a);
        }

        public void print(String a) {
            pw.print(a);
        }

        public String readLine() throws IOException {
            byte[] buf = new byte[3000064]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public void sort(int[] arr) {
            ArrayList<Integer> arlist = new ArrayList<>();
            for (int i : arr)
                arlist.add(i);

            Collections.sort(arlist);
            for (int i = 0; i < arr.length; i++)
                arr[i] = arlist.get(i);
        }

        public void sort(long[] arr) {
            ArrayList<Long> arlist = new ArrayList<>();
            for (long i : arr)
                arlist.add(i);

            Collections.sort(arlist);
            for (int i = 0; i < arr.length; i++)
                arr[i] = arlist.get(i);
        }

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

            return arr;
        }

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

            return arr;
        }

        public double[] nextDoubleArray(int n) throws IOException {
            double[] arr = new double[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextDouble();
            }

            return arr;
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            /*if (din == null)
                return;*/
            if (din != null) din.close();
            if (pw != null) pw.close();
        }
    }
}
