Problem: You are given n strings ti. Each string has cost ci.

Let's define the function of string <image>, where ps, i is the number of occurrences of s in ti, |s| is the length of the string s. Find the maximal value of function f(s) over all strings.

Note that the string s is not necessarily some string from t.

Input

The first line contains the only integer n (1 ≤ n ≤ 105) — the number of strings in t.

Each of the next n lines contains contains a non-empty string ti. ti contains only lowercase English letters.

It is guaranteed that the sum of lengths of all strings in t is not greater than 5·105.

The last line contains n integers ci ( - 107 ≤ ci ≤ 107) — the cost of the i-th string.

Output

Print the only integer a — the maximal value of the function f(s) over all strings s. Note one more time that the string s is not necessarily from t.

Examples

Input

2
aa
bb
2 1


Output

4


Input

2
aa
ab
2 1


Output

5

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class ExpensiveStrings {

	static int N,TN;
	static int[] dToDelim, sa, lcp, stringId;
	static long[] cost, prefix;
	static char c[], inp[][];
	static long res;
	static RMinQ rmq;
	public static void main(String[] args) {
		//Read in the input
		FS in = new FS();
		N = in.nextInt();
		inp = new char[N][];
		for(int i = 0; i < N; i++) {
			inp[i] = in.next().toCharArray();
			TN += (inp[i].length + 1);
		}
		cost = new long[N];
		for(int i = 0; i < N; i++) cost[i] = in.nextInt();
		
		c = new char[TN]; //Single string with all strings appended
		dToDelim = new int[TN]; //Distance to the delimiter
		stringId = new int[TN]; //Index of which input string
		int idd = 0;
		for(int i = 0; i < N; i++) {
			for(int j = 0; j < inp[i].length; j++) {
				dToDelim[idd] = inp[i].length - j;
				stringId[idd] = i;
				c[idd++] = inp[i][j];
			}
			c[idd++] = '~';
		}

		sa = SA(c); //Create suffix array
		
		//Create a prefix sum where value[i] = cost of string at i
		prefix = prefixSum(sa);
		
		//Create the lcp array and make sure nothing matches past the delimiter 
		lcp = lcp(sa, c);
		for(int i = 0; i < TN-1; i++) {
			lcp[i] = Math.min(lcp[i], Math.min(dToDelim[sa[i]], dToDelim[sa[i+1]]));
		}
		rmq = new RMinQ(lcp); //RMQ (min query) of the lcp array

		//Initialize res to be 0. If all possible subtrings give a negative
		// value, then you can choose some string that is not a substring
		// of anything, giving a value of 0
		res = 0;
		
		//Go forward in the lcp array, then reverse the arrays and
		// do the same thing again
// 		for(int iter = 0; iter < 2; iter++) {
			for(int i = 0; i < TN-1; i++) {
				if(dToDelim[sa[i]] == 0) continue; //delimiter
				//We might want to pick the whole suffix and not match
				// any other strings. This only works if our length
				// is larger than both the LCP to the left and right
				int lengthOfSuffix = dToDelim[sa[i]];
				int maxOfLcp = Math.max(lcp[i], i == 0 ? 0 : lcp[i-1]);
				if(lengthOfSuffix > maxOfLcp) {
					res = Math.max(res, lengthOfSuffix * cost[stringId[sa[i]]]);
				}
				
				//Our string length is the LCP of us and what is in front of us
				//Binary search how much this string matches with everything else
				// and then get the sum of values in that range
				int len = lcp[i];
				if(len == 0) continue; //no match
				//Get how far we match to the left
				int l = bsL(0, i, len);
				//Get how far we match to the right
				int r = bsR(i,TN-1,len);
				//The score for this string is equal to |S| * sum[l,r]
				long curRes = len * sum(l,r);
				res = Math.max(res, curRes);
			}
		
			//Reverse the suffix array and corresponding values
// 			sa = reverse(sa);
// 			lcp = reverse(lcp);
// 			prefix = prefixSum(sa);
// 			rmq = new RMinQ(lcp);
// 		}
		
		System.out.println(res);
	}

	//Returns the min lcp matching from s[l] to s[r] inclusive 
	static int lcpFrom(int l, int r) {
		if(l == r) return TN+1; //A string matches all the way with itself
		return rmq.query(l, r-1);
	}
	
	//Binary search the furthest left string such that
	// lcp from X to r is >= len
	static int bsL(int l, int r, int len) {
		int R = r;
		while(l < r) {
			int m = (l+r)>>1;
			if(lcpFrom(m,R) >= len) r = m;
			else l = m+1;
		}
		return l;
	}
	//Binary search the furthest right string such that
	// lcp from l to X is >= len
	static int bsR(int l, int r, int len) {
		int L = l;
		while(l < r) {
			int m = (l+r+1)>>1;
			if(lcpFrom(L,m) >= len) l = m;
			else r = m-1;
		}
		return l;
	}
	
	//Constructs a prefix sum of costs given the suffix array
	static long[] prefixSum(int sa[]) {
		long res[] = new long[sa.length];
		for(int i = 0; i < sa.length; i++) {
			res[i] = cost[stringId[sa[i]]];
			if(i > 0) res[i] += res[i-1];
		}
		return res;
	}
	
	//Return the sum of [l,r]
	static long sum(int l, int r) {
		return prefix[r] - (l == 0 ? 0 : prefix[l-1]);
	}
	
	//Reverse an array
	static int[] reverse(int a[]) {
		int b[] = new int[a.length];
		for(int i = a.length-1, id = 0; i >= 0; i--, id++) b[id]=a[i];
		return b;
	}
	
	
	//Hackpack under here
	static int[] SA(char S[]) {
		int n = S.length;Integer[] order = new Integer[n];
		for(int i = 0; i < n; i++)
			order[i] = n-1-i;
		Arrays.sort(order, (a,b) -> Character.compare(S[a], S[b]));
		int[] sa = new int[n];
		int[] classes = new int[n];
		for(int i = 0; i < n; i++) {
			sa[i] = order[i];
			classes[i] = S[i];
		}
		for(int len = 1; len < n; len *= 2) {
			int[] c = classes.clone();
			for(int i = 0; i < n; i++) {
				classes[sa[i]] = i > 0 && c[sa[i-1]] == c[sa[i]] && sa[i-1] + len<n && c[sa[i-1] + len/2] == c[sa[i] + len/2] ? classes[sa[i-1]] : i;
			}
			int[] cnt = new int[n];
			for(int i = 0; i < n; i++) cnt[i] = i;
			int[] s = sa.clone();
			for(int i = 0; i < n; i++) {
				int s1 = s[i] - len;
				if(s1 >= 0)sa[cnt[classes[s1]]++] = s1;
			}
		}
		return sa;
	}
	static int[] lcp(int[] sa, char[]s) {
		int n = sa.length;
		int[] rank = new int[n];
		for(int i = 0; i < n; i++)
			rank[sa[i]] = i;
		int[] lcp = new int[n-1];
		for(int i = 0, h = 0; i < n; i++) {
			if(rank[i] < n-1) {
				for(int j = sa[rank[i]+1]; Math.max(i, j) + h < s.length && s[i+h] == s[j+h]; ++h);
				lcp[rank[i]] = h;
				if(h > 0) --h;
			}
		}
		return lcp;
	}
	
	static class RMinQ {
		int[] vs;
		int[][] lift;
		public RMinQ(int[] vs) {
			this.vs = vs;
			int n = vs.length;
			int maxlog = Integer.numberOfTrailingZeros(Integer.highestOneBit(n)) + 2;
			lift = new int[maxlog][n];
			for(int i = 0; i < n; i++)
				lift[0][i] = vs[i];
			int lastRange = 1;
			for(int lg=  1; lg < maxlog; lg++) {
				for(int i = 0; i < n; i++) {
					lift[lg][i] = Math.min(lift[lg-1][i], lift[lg-1][Math.min(i+lastRange, n-1)]);
				}
				lastRange *= 2;
			}
		}
		public int query(int low, int hi) {
			int range = hi-low+1;
			int exp = Integer.highestOneBit(range);
			int lg = Integer.numberOfTrailingZeros(exp);
			return Math.min(lift[lg][low], lift[lg][hi-exp+1]);
		}
	}
	
	static class FS{
		BufferedReader br;
		StringTokenizer st;
		public FS() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch(Exception e) {throw null;}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
	}
	
}