Problem: Dima took up the biology of bacteria, as a result of his experiments, he invented k types of bacteria. Overall, there are n bacteria at his laboratory right now, and the number of bacteria of type i equals ci. For convenience, we will assume that all the bacteria are numbered from 1 to n. The bacteria of type ci are numbered from <image> to <image>.

With the help of special equipment Dima can move energy from some bacteria into some other one. Of course, the use of such equipment is not free. Dima knows m ways to move energy from some bacteria to another one. The way with number i can be described with integers ui, vi and xi mean that this way allows moving energy from bacteria with number ui to bacteria with number vi or vice versa for xi dollars.

Dima's Chef (Inna) calls the type-distribution correct if there is a way (may be non-direct) to move energy from any bacteria of the particular type to any other bacteria of the same type (between any two bacteria of the same type) for zero cost.

As for correct type-distribution the cost of moving the energy depends only on the types of bacteria help Inna to determine is the type-distribution correct? If it is, print the matrix d with size k × k. Cell d[i][j] of this matrix must be equal to the minimal possible cost of energy-moving from bacteria with type i to bacteria with type j.

Input

The first line contains three integers n, m, k (1 ≤ n ≤ 105; 0 ≤ m ≤ 105; 1 ≤ k ≤ 500). The next line contains k integers c1, c2, ..., ck (1 ≤ ci ≤ n). Each of the next m lines contains three integers ui, vi, xi (1 ≤ ui, vi ≤ 105; 0 ≤ xi ≤ 104). It is guaranteed that <image>.

Output

If Dima's type-distribution is correct, print string «Yes», and then k lines: in the i-th line print integers d[i][1], d[i][2], ..., d[i][k] (d[i][i] = 0). If there is no way to move energy from bacteria i to bacteria j appropriate d[i][j] must equal to -1. If the type-distribution isn't correct print «No».

Examples

Input

4 4 2
1 3
2 3 0
3 4 0
2 4 1
2 1 2


Output

Yes
0 2
2 0


Input

3 1 2
2 1
1 2 0


Output

Yes
0 -1
-1 0


Input

3 2 2
2 1
1 2 0
2 3 1


Output

Yes
0 1
1 0


Input

3 0 2
1 2


Output

No

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class D {

	class Node {
		int i;
		boolean visited = false;
		ArrayList<Node> nodes = new ArrayList<Node>();
		public Node(int i) {
			this.i = i;
		}
	}
	
	void solve() throws Exception {
		
		int n = nextInt();
		int m = nextInt();
		int k = nextInt();
		int[] c = nextInts(k);
		
		
		int[][] d = new int[k][k];
		for (int i=0; i<k; i++) for (int j=0; j<k; j++) d[i][j] = Integer.MAX_VALUE;
		for (int i=0; i<k; i++) d[i][i] = 0;
		
		// compute type of each bacteria
		int[] type = new int[n];
		int r = 0;
		for (int i=0; i<n; i++) {
			if (c[r] == 0) r++;
			c[r]--;
			type[i] = r;
		}
		
		Node[] nodes = new Node[n];
		for (int i=0; i<n; i++) nodes[i] = new Node(i);
		
		for (int i=0; i<m; i++) {
			int u = nextInt()-1;
			int v = nextInt()-1;
			int x = nextInt();
			
			int tu = type[u];
			int tv = type[v];
			
//			System.out.println("From type " + tu + " to " + tv + " for " + x);
			
//			if (d[tu][tv] != Integer.MAX_VALUE && d[tu][tv] != x) {
//				System.out.println("No");
//				return;
//			}
			
			if (x == 0) {
				Node un = nodes[u];
				Node vn = nodes[v];
				un.nodes.add(vn);
				vn.nodes.add(un);
			}
			
			d[tu][tv] = d[tv][tu] = Math.min(d[tu][tv], x);
		}
		
			
		// bfs from first node of each type to determine reachability
		int t = -1;
		for (int i=0; i<n; i++) {
			Node start = nodes[i];
			if (type[i] == t) {
				if (!start.visited) {
					System.out.println("No");
					return;
				}
				continue;
			}
			t = type[i];
			
			// bfs
			ArrayDeque<Node> q = new ArrayDeque<Node>();
			q.add(start);
			while(!q.isEmpty()) {
				Node node = q.poll();
				if (node.visited) continue;
				node.visited = true;
				for (Node node2 : node.nodes) {
					if (node2.visited) continue;
					q.add(node2);
				}
			}
			
		}
		
		
		
		
		// floyd warshall to find shortest path for the groups
		for (int f=0; f<k; f++)
			for (int i=0; i<k; i++) {
				if (d[i][f] == Integer.MAX_VALUE) continue;
				for (int j=0; j<k; j++) {
					if (d[f][j] == Integer.MAX_VALUE) continue;
					int a = d[i][f] + d[f][j];
					if (a >= d[i][j]) continue;
					d[i][j] = a;
				}
			}
		
		System.out.println("Yes");
		StringBuilder sb = new StringBuilder();
		for (int i=0; i<k; i++) {
			for (int j=0; j<k; j++) {
				if (j != 0) sb.append(" ");
				if (d[i][j] == Integer.MAX_VALUE) sb.append(-1);
				else sb.append(d[i][j]);
			}
			sb.append("\n");
		}
		System.out.print(sb);
		
	}

	
	
	
	
	
	
	
	
	
	
	
	
	
	BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st = new StringTokenizer("");

	String nextLine() throws Exception {
		String line = stdin.readLine();
		st = new StringTokenizer("");
		return line;
	}
	int[] nextInts(int n) throws Exception {
		int[] res = new int[n];
		for (int i=0; i<n; i++) res[i] = nextInt();
		return res;
	}
	long[] nextLongs(int n) throws Exception {
		long[] res = new long[n];
		for (int i=0; i<n; i++) res[i] = nextLong();
		return res;
	}
	double[] nextDoubles(int n) throws Exception {
		double[] res = new double[n];
		for (int i=0; i<n; i++) res[i] = nextDouble();
		return res;
	}
	String next() throws Exception {
		while (!st.hasMoreTokens()) 
			st = new StringTokenizer(stdin.readLine());
		return st.nextToken();
	}
	int nextInt() throws Exception {
		return Integer.parseInt(next());
	}
	long nextLong() throws Exception {
		return Long.parseLong(next());
	}
	double nextDouble() throws Exception {
		return Double.parseDouble(next());
	}
	public static void main(String[] args) throws Exception {
		new D().solve();
	}
}