Problem:
Score : 400 points

Problem Statement

A university student, Takahashi, has to take N examinations and pass all of them.
Currently, his readiness for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination.

Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness.

For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions:

The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal.

For every i, B_i \leq C_i holds.

If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

A_i and B_i are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{N}
B_1 B_2 ... B_{N}

Output

Print the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the conditions.
If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.

Sample Input 1

3
2 3 5
3 4 1

Sample Output 1

3

(A_1, A_2, A_3) = (2, 3, 5) and (B_1, B_2, B_3) = (3, 4, 1). If nothing is done, he cannot pass the first and second exams.
The minimum possible number of indices i such that A_i and C_i are different, 3, is achieved when:

(C_1, C_2, C_3) = (3, 5, 2)

Sample Input 2

3
2 3 3
2 2 1

Sample Output 2

0

In this case, he has to do nothing in order to pass all the exams.

Sample Input 3

3
17 7 1
25 6 14

Sample Output 3

-1

In this case, no matter what is done, he cannot pass all the exams.

Sample Input 4

12
757232153 372327760 440075441 195848680 354974235 458054863 463477172 740174259 615762794 632963102 529866931 64991604
74164189 98239366 465611891 362739947 147060907 118867039 63189252 78303147 501410831 110823640 122948912 572905212

Sample Output 4

5

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.Collections;
import java.util.NoSuchElementException;

public class Main {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int N = sc.nextInt();
		long[] A = new long[N];
		for (int i = 0; i < N; i++)
			A[i] = sc.nextInt();
		long[] B = new long[N];
		for (int i = 0; i < N; i++)
			B[i] = sc.nextInt();

		long sa = 0;
		for (int i = 0; i < N; i++)
			sa += A[i];

		long sb = 0;
		for (int i = 0; i < N; i++)
			sb += B[i];

		if (sa < sb) {
			System.out.println("-1");
			return;
		}
		int count = 0;
		long s = 0;
		Long[] D = new Long[N];
		for (int i = 0; i < N; i++)
			D[i] = 0L;

		for (int i = 0; i < N; i++) {
			if (A[i] < B[i]) {
				s += B[i] - A[i];
				count++;
			} else if (A[i] >= B[i]) {
				D[i] = A[i] - B[i];
			}
		}
		Arrays.sort(D, Collections.reverseOrder());
		int i = 0;
		while (s > 0) {
			s -= D[i];
			count++;
			i++;
		}
		System.out.println(count);

	}

	//以下、自作ライブラリ

	/*
		int N = sc.nextInt();

		int[] a = new int[N];
		for (int i = 0; i < N; i++)
			a[i] = sc.nextInt();

		String s = sc.next();

		String[] a = new String[N];
		for (int i = 0; i < N; i++)
			a[i] = sc.next();

		int[][] a = new int[N][2];
		for (int i = 0; i < N; i++) {
			a[i][0] = sc.nextInt();
			a[i][1] = sc.nextInt();
		}
		Arrays.sort(a, (x, y) -> Integer.compare(x[0], y[0]));

		List<Integer> list = new ArrayList<>();

		Set<Integer> set = new HashSet<>();

		Map<Integer, Integer> map = new HashMap<>();
	 */

	//階乗
	public static int fact(int n) {
		if (n == 1 || n == 0)
			return 1;
		else
			return n * fact(n - 1);
	}

	//指定した値以上の要素が最初に現れる位置
	//要素の数は+1（配列のズレ)
	public static long lowerBound(int n, long[] x, long y) {
		int left = 0;
		int right = n;
		while (left < right) {
			int mid = (left + right) / 2;
			if (x[mid] < y) {
				left = mid + 1;
			} else {
				right = mid;
			}
		}
		return left;
	}

	//指定した値より大きい要素が最初に現れる位置
	//要素の数は+1（配列のズレ)
	public static long upperBound(int n, long[] x, long y) {
		int left = 0;
		int right = n;
		while (left < right) {
			int mid = (left + right) / 2;
			if (x[mid] <= y) {
				left = mid + 1;
			} else {
				right = mid;
			}
		}
		return left;
	}

	//最大公約数
	public static long gcd(long a, long b) {
		if (a < b)
			return gcd(b, a);
		long c = 0;
		c = a % b;
		if (c == 0)
			return b;
		return gcd(b, c);
	}

	//最小公倍数
	public static long lcm(long m, long n) {
		return m * n / gcd(m, n);
	}

	//素数判定
	public static boolean isPrime(int n) {
		if (n < 2)
			return false;
		if (n == 2)
			return true;
		if (n % 2 == 0)
			return false;

		for (int i = 3; i * i <= n; i += 2) {
			if (n % i == 0)
				return false;
		}
		return true;
	}
}

//以下、高速スキャナークラス
class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
		return hasNextByte();
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
			throw new NumberFormatException();
		return (int) nl;
	}

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