Problem: Nastya just made a huge mistake and dropped a whole package of rice on the floor. Mom will come soon. If she sees this, then Nastya will be punished.

In total, Nastya dropped n grains. Nastya read that each grain weighs some integer number of grams from a - b to a + b, inclusive (numbers a and b are known), and the whole package of n grains weighs from c - d to c + d grams, inclusive (numbers c and d are known). The weight of the package is the sum of the weights of all n grains in it.

Help Nastya understand if this information can be correct. In other words, check whether each grain can have such a mass that the i-th grain weighs some integer number x_i (a - b ≤ x_i ≤ a + b), and in total they weigh from c - d to c + d, inclusive (c - d ≤ ∑_{i=1}^{n}{x_i} ≤ c + d).

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. 

The next t lines contain descriptions of the test cases, each line contains 5 integers: n (1 ≤ n ≤ 1000) — the number of grains that Nastya counted and a, b, c, d (0 ≤ b < a ≤ 1000, 0 ≤ d < c ≤ 1000) — numbers that determine the possible weight of one grain of rice (from a - b to a + b) and the possible total weight of the package (from c - d to c + d).

Output

For each test case given in the input print "Yes", if the information about the weights is not inconsistent, and print "No" if n grains with masses from a - b to a + b cannot make a package with a total mass from c - d to c + d.

Example

Input


5
7 20 3 101 18
11 11 10 234 2
8 9 7 250 122
19 41 21 321 10
3 10 8 6 1


Output


Yes
No
Yes
No
Yes

Note

In the first test case of the example, we can assume that each grain weighs 17 grams, and a pack 119 grams, then really Nastya could collect the whole pack.

In the third test case of the example, we can assume that each grain weighs 16 grams, and a pack 128 grams, then really Nastya could collect the whole pack.

In the fifth test case of the example, we can be assumed that 3 grains of rice weigh 2, 2, and 3 grams, and a pack is 7 grams, then really Nastya could collect the whole pack.

In the second and fourth test cases of the example, we can prove that it is impossible to determine the correct weight of all grains of rice and the weight of the pack so that the weight of the pack is equal to the total weight of all collected grains.

Solution (in PYTHON3):

Solution:
## necessary imports
import sys
input = sys.stdin.readline
from math import log2, log, ceil

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b%a, a)

## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k 
    res = 1
    for i in range(k): 
        res = res * (n - i) 
        res = res / (i + 1) 
    return res 

## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
    hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x:
            lo = mid+1
        else:
            hi = mid
    return lo

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function
def find(x, link):
    while(x != link[x]):
        x = link[x]
    return x

# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, size, link):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)] 
# spf_sieve() 
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().strip().split()))
## taking string array input
def str_array():
    return input().strip().split();

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');


################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

for _ in range(int(input())):
    n, a, b, c, d = int_array();
    if min(n*(a+b), c+d) - max(n*(a-b), c-d) >= 0:
        print('Yes')
    else:
        print('No');