You are given two arrays A and B, each of size n. The error, E, between these two arrays is defined <image>. You have to perform exactly k1 operations on array A and exactly k2 operations on array B. In one operation, you have to choose one element of the array and increase or decrease it by 1.

Output the minimum possible value of error after k1 operations on array A and k2 operations on array B have been performed.

Input

The first line contains three space-separated integers n (1 ≤ n ≤ 103), k1 and k2 (0 ≤ k1 + k2 ≤ 103, k1 and k2 are non-negative) — size of arrays and number of operations to perform on A and B respectively.

Second line contains n space separated integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — array A.

Third line contains n space separated integers b1, b2, ..., bn ( - 106 ≤ bi ≤ 106)— array B.

Output

Output a single integer — the minimum possible value of <image> after doing exactly k1 operations on array A and exactly k2 operations on array B.

Examples

Input

2 0 0
1 2
2 3


Output

2

Input

2 1 0
1 2
2 2


Output

0

Input

2 5 7
3 4
14 4


Output

1

Note

In the first sample case, we cannot perform any operations on A or B. Therefore the minimum possible error E = (1 - 2)2 + (2 - 3)2 = 2. 

In the second sample case, we are required to perform exactly one operation on A. In order to minimize error, we increment the first element of A by 1. Now, A = [2, 2]. The error is now E = (2 - 2)2 + (2 - 2)2 = 0. This is the minimum possible error obtainable.

In the third sample case, we can increase the first element of A to 8, using the all of the 5 moves available to us. Also, the first element of B can be reduced to 8 using the 6 of the 7 available moves. Now A = [8, 4] and B = [8, 4]. The error is now E = (8 - 8)2 + (4 - 4)2 = 0, but we are still left with 1 move for array B. Increasing the second element of B to 5 using the left move, we get B = [8, 5] and E = (8 - 8)2 + (4 - 5)2 = 1.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.*;

public class Minimizetheerror {

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			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;
		}
	}

	static class Error {
		long error;
		int index;

		@Override
		public String toString() {
			return "Error [error=" + error + ", index=" + index + "]";
		}

		public Error(long error, int index) {
			super();
			this.error = error;
			this.index = index;
		}

	}

	public static void main(String[] args) {

		FastScanner f = new FastScanner();
		PrintWriter p = new PrintWriter(System.out);
		int n = f.nextInt();
		long a = f.nextLong();
		long b = f.nextLong();
		long arr1[] = new long[n];
		long arr2[] = new long[n];
		for (int i = 0; i < n; i++) {
			arr1[i] = f.nextLong();
		}
		for (int i = 0; i < n; i++) {
			arr2[i] = f.nextLong();
		}

		PriorityQueue<Long> pq = new PriorityQueue<>(Collections.reverseOrder());
		long k = a + b;

		for (int i = 0; i < n; i++) {
			pq.add(Math.abs(arr1[i] - arr2[i]));
		}

		while (k > 0) {
			long curr = pq.peek();
			pq.poll();
			pq.add(Math.abs(curr - 1));
			k--;
		}
		long ans = 0;
		while (pq.isEmpty() == false) {
			long curr = pq.peek();
			ans += curr * curr;
			pq.poll();
		}

		p.println(ans);
		p.close();

	}

	/// Ceil
	static long ceil(long x, long m) {
		long res = x / m;
		if (x % m != 0) {
			res++;
		}
		return res;
	}

	// ------------------------------------------------------------------------------------------------
	// makes the prefix sum array
	static long[] prefixSum(long arr[], int n) {
		long psum[] = new long[n];
		psum[0] = arr[0];
		for (int i = 1; i < n; i++) {
			psum[i] = psum[i - 1] + arr[i];
		}
		return psum;
	}
	// ------------------------------------------------------------------------------------------------

	// makes the suffix sum array
	static long[] suffixSum(long arr[], int n) {
		long ssum[] = new long[n];
		ssum[n - 1] = arr[n - 1];
		for (int i = n - 2; i >= 0; i--) {
			ssum[i] = ssum[i + 1] + arr[i];
		}
		return ssum;
	}

//------------------------------------------------------------------------------------------
	// BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
	// OVERHEADS
	static long m = (long) (1e9 + 7);

	static long binPower(long a, long n, long m) {
		if (n == 0)
			return 1;
		long res = 1;
		while (n > 0) {
			if ((n & 1) != 0) {
				res *= a;
			}
			a *= a;
			n >>= 1;
		}
		return res;
	}

//-------------------------------------------------------------------------------------------

	// gcd
	static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

//------------------------------------------------------------------------------------------

	// lcm
	static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}

//------------------------------------------------------------------------------------------
	// BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
	// O(LOGn)
	static int setBits(int n) {
		int count = 0;
		while (n > 0) {
			n = n & (n - 1);
			count++;
		}
		return count;
	}
//------------------------------------------------------------------------------------------

//------------------------------------------------------------------------------------------
	// 0 based indexing
	static boolean KthBitSet(int n, int k) {
		int mask = 1;
		mask = mask <<= k;
		if ((mask & n) != 0)
			return true;
		else
			return false;
	}

//------------------------------------------------------------------------------------------

	// EXTENDED EUCLIDEAN THEOREM
	// TO REPRESENT GCD IN TERMS OF A AND B
	// gcd(a,b) = a.x + b.y where x and y are integers

	static long x = -1;
	static long y = -1;

	static long gcdxy(long a, long b) {
		if (b == 0) {
			x = 1;
			y = 0;
			return a;
		} else {
			long d = gcdxy(b, a % b);
			long x1 = y;
			long y1 = x - (a / b) * y;
			x = x1;
			y = y1;

			return d;
		}
	}

//-------------------------------------------------------------------------------------------------F
}
