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.
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
import java.lang.*;
import java.util.function.Predicate;

public class Main{
    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static long MOD = (long) (1e9 + 7);
    //    static long MOD = 998244353;
    static long MOD2 = MOD * MOD;
    static FastReader sc = new FastReader();
    static int pInf = Integer.MAX_VALUE;
    static int nInf = Integer.MIN_VALUE;
    static long ded = (long)(1e17)+9;
    public static void main(String[] args) throws Exception {
        int test = 1;
//        test = sc.nextInt();
        for (int i = 1; i <= test; i++){
//            out.print("Case #"+i+": ");
            solve();
        }
        out.flush();
        out.close();
    }
    static int n,m;
    static int[][] hor,ver;
    static Long[][][] dp;
    static void solve(){
        n = sc.nextInt();
        m = sc.nextInt();
        int k = sc.nextInt();
        dp = new Long[n+1][m+1][k+1];
        hor = new int[n][m-1];
        ver = new int[n-1][m];
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m-1; j++){
                hor[i][j] = sc.nextInt();
            }
        }
        for(int i = 0; i < n-1; i++){
            for(int j = 0; j < m; j++){
                ver[i][j] = sc.nextInt();
            }
        }
        if(k%2==1){
            for(int i = 0; i < n; i++){
                for(int j = 0; j < m; j++){
                    out.print(-1+" ");
                }
                out.println();
            }
            return;
        }
        k = k/2;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                long ans = cal(i,j,k);
                out.print((2*ans)+" ");
            }
            out.println();
        }
    }
    static long cal(int i, int j,int k){
        if(k==0)return 0;
        if(dp[i][j][k]!=null)return dp[i][j][k];
        long ans = ded;
        for(int h = 0; h < 4; h++){
            int ni = i+di[h];
            int nj = j+dj[h];
            if(e(ni,nj)){
                int cost = 0;
                if(ni==i){
                    if(nj>j){
                        cost = hor[i][j];
                    }else{
                        cost = hor[i][nj];
                    }
                }if(nj==j){
                    if(ni>i){
                        cost = ver[i][j];
                    }else{
                        cost = ver[ni][j];
                    }
                }
                ans = Math.min(ans,(cost)+cal(ni,nj,k-1));
            }
        }
        return dp[i][j][k] =  ans;
    }
    static int[] di = new int[]{0,-1,0,1};
    static int[] dj = new int[]{-1,0,1,0};
    static boolean e(int i, int j){
        return i>=0&&j>=0&&i<n&&j<m;
    }
    static class Pair implements Comparable<Pair> {
        int x;
        int y;
        public Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
        @Override
        public int compareTo(Pair o){
            return this.x-o.x;
        }

        @Override
        public String toString() {
            return "Pair{" + "x=" + x + ", y=" + y + '}';
        }

        public boolean equals(Pair o){
            return this.x==o.x&&this.y==o.y;
        }
    }
    public static long mul(long a, long b) {
        return ((a % MOD) * (b % MOD)) % MOD;
    }
    public static long add(long a, long b) {
        return ((a % MOD) + (b % MOD)) % MOD;
    }
    public static long c2(long n) {
        if ((n & 1) == 0) {
            return mul(n / 2, n - 1);
        } else {
            return mul(n, (n - 1) / 2);
        }
    }
    //Shuffle Sort
    static final Random random = new Random();
    static void ruffleSort(int[] a) {
        int n = a.length;//shuffle, then sort
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n); int temp= a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }
    //Brian Kernighans Algorithm
    static long countSetBits(long n) {
        if (n == 0) return 0;
        return 1 + countSetBits(n & (n - 1));
    }
    //Euclidean Algorithm
    static long gcd(long A, long B) {
        if (B == 0) return A;
        return gcd(B, A % B);
    }
    //Modular Exponentiation
    static long fastExpo(long x, long n) {
        if (n == 0) return 1;
        if ((n & 1) == 0) return fastExpo((x * x) % MOD, n / 2) % MOD;
        return ((x % MOD) * fastExpo((x * x) % MOD, (n - 1) / 2)) % MOD;
    }
    //AKS Algorithm
    static boolean isPrime(long n) {
        if (n <= 1) return false;
        if (n <= 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        for (int i = 5; i <= Math.sqrt(n); i += 6)
            if (n % i == 0 || n % (i + 2) == 0) return false;
        return true;
    }
    public static long modinv(long x) {
        return modpow(x, MOD - 2);
    }
    public static long modpow(long a, long b) {
        if (b == 0) {
            return 1;
        }
        long x = modpow(a, b / 2);
        x = (x * x) % MOD;
        if (b % 2 == 1) {
            return (x * a) % MOD;
        }
        return x;
    }


    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}