Problem:
Score : 400 points

Problem Statement

There are N squares in a row.
The leftmost square contains the integer A, and the rightmost contains the integer B. The other squares are empty.

Aohashi would like to fill the empty squares with integers so that the following condition is satisfied:

For any two adjacent squares, the (absolute) difference of the two integers in those squares is between C and D (inclusive).

As long as the condition is satisfied, it is allowed to use arbitrarily large or small integers to fill the squares.
Determine whether it is possible to fill the squares under the condition.

Constraints

3 \leq N \leq 500000

0 \leq A \leq 10^9

0 \leq B \leq 10^9

0 \leq C \leq D \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B C D

Output

Print YES if it is possible to fill the squares under the condition; print NO otherwise.

Sample Input 1

5 1 5 2 4

Sample Output 1

YES

For example, fill the squares with the following integers: 1, -1, 3, 7, 5, from left to right.

Sample Input 2

4 7 6 4 5

Sample Output 2

NO

Sample Input 3

48792 105960835 681218449 90629745 90632170

Sample Output 3

NO

Sample Input 4

491995 412925347 825318103 59999126 59999339

Sample Output 4

YES

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    static int N;
    static int A;
    static int B;
    static int C;
    static int D;

    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        N = sc.nextInt();
        A = sc.nextInt();
        B = sc.nextInt();
        C = sc.nextInt();
        D = sc.nextInt();

        System.out.println( solve() ? "YES" : "NO" );
    }

    static boolean solve() {
        // GはgoalのG
        int G = Math.abs(A-B);
        int times = N - 1;

        // C*q D*q G という状態まで詰める
        int q = G / D;
        int r = G % D;
        times -= q;

        // 回数切れで届かない
        if( times < 0 ) return false;
        // ぴったりのケース(ここでやらなくてもよい気がする)
        if( times == 0 ) return r == 0;

        if( times % 2 == 0 ) {
            // +D -Cとすることで D-Cずつ距離を詰めることができる
            long available = (times / 2) * (long)(D-C);
            return r <= available;

        } else {
            // もう一歩進む
            // G C1 D1 or C1 G D1
            long C1 = C * (q+1);
            long D1 = D * (q+1);
            times -= 1;

            // C1 <= G <= D1 の場合はそれで終了(0で足踏みすればよい)
            if( C1 <= G && G <= D1 ) return true;

            // G D1 C1 という状態なので D1からGに向かって詰めることができたらOK
            long r1 = D1 - G;
            long available = (times / 2) * (long)(D-C);
            return r1 <= available;
        }
    }

    static class FastScanner {

        private BufferedReader reader;
        private StringTokenizer tokenizer;

        FastScanner(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        String next() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }

            return tokenizer.nextToken("\n");
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

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

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
    }
}