Problem: The farmer Polycarp has a warehouse with hay, which can be represented as an n × m rectangular table, where n is the number of rows, and m is the number of columns in the table. Each cell of the table contains a haystack. The height in meters of the hay located in the i-th row and the j-th column is equal to an integer ai, j and coincides with the number of cubic meters of hay in the haystack, because all cells have the size of the base 1 × 1. Polycarp has decided to tidy up in the warehouse by removing an arbitrary integer amount of cubic meters of hay from the top of each stack. You can take different amounts of hay from different haystacks. Besides, it is allowed not to touch a stack at all, or, on the contrary, to remove it completely. If a stack is completely removed, the corresponding cell becomes empty and no longer contains the stack.

Polycarp wants the following requirements to hold after the reorganization:

  * the total amount of hay remaining in the warehouse must be equal to k, 
  * the heights of all stacks (i.e., cells containing a non-zero amount of hay) should be the same, 
  * the height of at least one stack must remain the same as it was, 
  * for the stability of the remaining structure all the stacks should form one connected region. 



The two stacks are considered adjacent if they share a side in the table. The area is called connected if from any of the stack in the area you can get to any other stack in this area, moving only to adjacent stacks. In this case two adjacent stacks necessarily belong to the same area.

Help Polycarp complete this challenging task or inform that it is impossible.

Input

The first line of the input contains three integers n, m (1 ≤ n, m ≤ 1000) and k (1 ≤ k ≤ 1018) — the number of rows and columns of the rectangular table where heaps of hay are lain and the required total number cubic meters of hay after the reorganization. 

Then n lines follow, each containing m positive integers ai, j (1 ≤ ai, j ≤ 109), where ai, j is equal to the number of cubic meters of hay making the hay stack on the i-th row and j-th column of the table.

Output

In the first line print "YES" (without quotes), if Polycarpus can perform the reorganisation and "NO" (without quotes) otherwise. If the answer is "YES" (without quotes), then in next n lines print m numbers — the heights of the remaining hay stacks. All the remaining non-zero values should be equal, represent a connected area and at least one of these values shouldn't be altered.

If there are multiple answers, print any of them.

Examples

Input

2 3 35
10 4 9
9 9 7


Output

YES
7 0 7 
7 7 7 


Input

4 4 50
5 9 1 1
5 1 1 5
5 1 5 5
5 5 7 1


Output

YES
5 5 0 0 
5 0 0 5 
5 0 5 5 
5 5 5 0 


Input

2 4 12
1 1 3 1
1 6 2 4


Output

NO

Note

In the first sample non-zero values make up a connected area, their values do not exceed the initial heights of hay stacks. All the non-zero values equal 7, and their number is 5, so the total volume of the remaining hay equals the required value k = 7·5 = 35. At that the stack that is on the second line and third row remained unaltered.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class F {

	final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
	BufferedReader in;
	PrintWriter out;
	StringTokenizer tok = new StringTokenizer("");

	void solve() throws IOException {
		int t = 1;
		while (t-- > 0) {
			solveTest();
		}
	}

	int n;
	int m;
	int[][] g;

	List<Integer> cur;

	boolean[][] used;

	class DSU {
		int[] p;
		int[] size;
		int n;
		Random rg = new Random();
		int maxSize = 1;
		int maxIndex = -1;

		DSU(int n) {
			p = new int[n];
			size = new int[n];
			this.n = n;
			for (int i = 0; i < n; i++) {
				p[i] = i;
				size[i] = 1;
			}
		}

		int getParent(int v) {
			if (p[v] == v) return v;
			return p[v] = getParent(p[v]);
		}

		void merge(int a, int b) {
			a = getParent(a);
			b = getParent(b);
			if (a == b) return;
			if (size[a] > size[b]) {
				size[a] += size[b];
				p[b] = a;
				if (size[a] > maxSize) {
					maxSize = size[a];
					maxIndex = a;
				}
			} else {
				size[b] += size[a];
				p[a] = b;
				if (size[b] > maxSize) {
					maxSize = size[b];
					maxIndex = b;
				}
			}
		}
	}

	class Stog implements Comparable<Stog> {
		int x;
		int y;
		int index;
		int value;

		public Stog(int x, int y, int value) {
			this.x = x;
			this.y = y;
			this.value = value;
			this.index = x * (m+2) + y;
		}

		@Override
		public int compareTo(Stog o) {
			return Integer.compare(this.value, o.value);
		}
	}

	void solveTest() throws IOException {
		n = readInt();
		m = readInt();
		long k = readLong();
		g = new int[n + 2][m + 2];

		Stog[] z = new Stog[n * m];
		int ptr = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				g[i][j] = readInt();
				z[ptr++] = new Stog(i, j, g[i][j]);
			}
		}
		used = new boolean[n + 2][m + 2];
		DSU dsu = new DSU((n + 2) * (m + 2));
		Arrays.sort(z);

		int i = n * m - 1;
		int resP = -1;
		int resS = -1;

		while (i >= 0) {
			int j = i;
			dsu.maxSize = 0;
			dsu.maxIndex = -1;
			while (j >= 0 && z[j].value == z[i].value) {
				Stog curStog = z[j];
				used[curStog.x][curStog.y] = true;

				for(int step = 0; step < 4; step++) {
					int nx = curStog.x + sx[step];
					int ny = curStog.y + sy[step];
					if(!used[nx][ny]) continue;
					dsu.merge(curStog.index, nx * (m + 2) + ny);
				}
				j--;
			}
			int max = dsu.maxSize;
			if(max == 0) {
				max = 1;
				dsu.maxIndex = z[i].index;
			}
			if(k % z[i].value != 0) {
				i = j;
				continue;
			}
			long need = k / z[i].value;
			if(need <= max) {
				resP = dsu.maxIndex;
				resS = z[i].value;
				break;
			}
			i = j;
		}
		if(resP == -1) {
			out.println("NO");
			return;
		}
		out.println("YES");

		long need = k / resS;
		int x = -1;
		int y = -1;
		for(i = 1; i <= n; i++) {
			for(int j = 1; j <= m; j++) {
				if(dsu.getParent(i * (m + 2) + j) == resP && g[i][j] == resS) {
					x = i;
					y = j;
					break;
				}
			}
		}

		res = new int[n+2][m+2];
		used = new boolean[n+2][m+2];
		count = need;
		dfs(x, y, resS);

		for(i = 1; i <= n; i++){
			for(int j = 1; j <= m; j++) {
				out.print(res[i][j] + " ");
			}
			out.println();
		}
	}
	long count;
	void dfs(int x, int y, int s) {
		used[x][y] = true;
		res[x][y] = s;
		count--;
		for(int i = 0; i < 4; i++) {
			int nx = x + sx[i];
			int ny = y + sy[i];
			if(g[nx][ny] < s || used[nx][ny]) continue;
			if(count == 0) continue;
			dfs(nx, ny, s);
		}
	}


	int[][] res;

	int[] sx = {-1, 0, 1, 0};
	int[] sy = {0, -1, 0, 1};

	void init() throws FileNotFoundException {
		if (ONLINE_JUDGE) {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
		} else {
			in = new BufferedReader(new FileReader("input.txt"));
			out = new PrintWriter("output.txt");
		}
	}

	String readString() throws IOException {
		while (!tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	int readInt() throws IOException {
		return Integer.parseInt(readString());
	}

	long readLong() throws IOException {
		return Long.parseLong(readString());
	}

	double readDouble() throws IOException {
		return Double.parseDouble(readString());
	}

	int[] readArr(int n) throws IOException {
		int[] res = new int[n];
		for (int i = 0; i < n; i++) {
			res[i] = readInt();
		}
		return res;
	}

	long[] readArrL(int n) throws IOException {
		long[] res = new long[n];
		for (int i = 0; i < n; i++) {
			res[i] = readLong();
		}
		return res;
	}

	public static void main(String[] args) {
		new F().run();
	}

	public void run() {
		try {
			long t1 = System.currentTimeMillis();
			init();
			solve();
			out.close();
			long t2 = System.currentTimeMillis();
			System.err.println("Time = " + (t2 - t1));
		} catch (Exception e) {
			e.printStackTrace(System.err);
			System.exit(-1);
		}
	}
}