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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define countof(a) (sizeof(a)/sizeof(*a))

#define vi vector<int>
#define vvi vector<vector<int> >
#define vpi vector<pi >
#define pi pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define all(n) n.begin(), n.end()

#define FROMTO(var, from, to) for (register int var = (from), var##down = ((int)(to)) < ((int)(from));var##down ? (var >= (int)(to)) : (var <= (int)(to));var##down ? var-- : var++)
#define UPTO(var, from, to)   for (register int var = (from); var <= ((int)to); var++)
#define DOWNTO(var, from, to) for (register int var = (from); var >= ((int)to); var--)
#define FOR(var, to)          UPTO(var, 0, (to)-1)
#define DOWN(var, from)       DOWNTO(var, (from)-1, 0) 

#define INIT(var, val) FOR(i,countof(var)) var[i] = val
#define INPUT(var) FOR(i,countof(var)) cin >> var[i]
#define INPUT1(var) FOR(i,countof(var)) cin >> var[i], var[i]--

#define SORT(v) qsort(v,countof(v),sizeof(*v),int_less)
#define SORTT(v) qsort(v,countof(v),sizeof(*v),int_greater)
#define QSORT(v,b) qsort(v,countof(v),sizeof(*v),b)

#define MOD 1000000007
#define INF ((1 << 30)-1)
#define LINF ((1LL << 62)-1)

typedef uint8_t u8;
typedef uint16_t u16;
typedef uint32_t u32;
typedef uint64_t u64;
typedef int8_t s8;
typedef int16_t s16;
typedef int32_t s32;
typedef int64_t s64;

struct Comb {
    vector<vector<s64> > data;
    Comb(int n) { // O(n^2)
        data = vector<vector<s64> >(n+1,vector<s64>(n+1,1));
        UPTO(i,1,n) {
            FOR(j,i+1) {
                if (!j || j == i) data[i][j] = 1;
                else data[i][j] = data[i-1][j-1] + data[i-1][j];
            }
        }
    }
    
    s64 ncr(int n, int r) {
        return data[n][r];
    }
};

static inline int ri() {
  int a;
  scanf("%d", &a);
  return a;
}

int int_less(const void *a, const void *b) {
  return (*((const int*)a) - *((const int*)b));
}
int int_greater(const void *a, const void *b) {
  return (*((const int*)b) - *((const int*)a));
}

int main() {
	int n = ri();
	int a = ri();
	int b = ri();
	int c = ri();
	int d = ri();
	
	bool res = false;
	FOR(i,n) { // plus
		int minus = n-1-i;
		s64 minn = a + ((s64)c)*i - ((s64)d)*minus;
		s64 maxx = a + ((s64)d)*i - ((s64)c)*minus;
		if (b >= minn && b <= maxx) res = true;
	}
	
	cout << (res?"YES":"NO") << endl;
	
    return 0;
}