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.
"""
    Satwik_Tiwari ;) .
    28 june , 2020  - Sunday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase


import bisect
from heapq import *
from math import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) : # Check Prime Number or not
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))
def bs(a,l,h,x):
    while(l<h):
        # print(l,h)
        mid = (l+h)//2
        if(a[mid] == x):
            return mid
        if(a[mid] < x):
            l = mid+1
        else:
            h = mid
    return l

def sieve(a): #O(n loglogn) nearly linear
    #all odd mark 1
    for i in range(3,((10**6)+1),2):
        a[i] = 1
    #marking multiples of i form i*i 0. they are nt prime
    for i in range(3,((10**6)+1),2):
        for j in range(i*i,((10**6)+1),i):
            a[j] = 0
    a[2] = 1 #special left case
    return (a)


def bfs(g,st):
    visited = [0]*(len(g))
    visited[st] = 1
    queue = []
    queue.append(st)
    new = []
    while(len(queue) != 0):
        s = queue.pop()
        new.append(s)
        for i in g[s]:
            if(visited[i] == 0):
                visited[i] = 1
                queue.append(i)
    return new

def chck(a):
    n = len(a)
    chnge= 0
    i = 1
    while(i<n):
        if(a[i] == a[i-1]):
            i+=1
            continue
        else:
            chnge+=1
            i+=1
    if(chnge == 2):
        if(a[0] == 'a'):
            return True
    if(chnge<2):
        return True
    return False


def solve():
    n,a,b,c,d = sep()
    # print(n*(a-b),n*(a+b),c-d,c+d)
    minn = n*(a-b)
    maxx = n*(a+b)

    if(c-d>maxx or c+d < minn):
        print('No')
    else:
        print('Yes')


# testcase(1)
testcase(int(inp()))








