You are given two integers l and r, l≤ r. Find the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.

As a reminder, a mod b is a remainder we get when dividing a by b. For example, 26 mod 8 = 2.

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1≤ t≤ 10^4), denoting the number of test cases. Description of the test cases follows.

The only line of each test case contains two integers l, r (1≤ l ≤ r ≤ 10^9).

Output

For every test case, output the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.

Example

Input


4
1 1
999999999 1000000000
8 26
1 999999999


Output


0
1
12
499999999

Note

In the first test case, the only allowed pair is (a, b) = (1, 1), for which a mod b = 1 mod 1 = 0.

In the second test case, the optimal choice is pair (a, b) = (1000000000, 999999999), for which a mod b = 1.
import sys,os,io
input = sys.stdin.readline # for strings
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline # for non-strings

PI = 3.141592653589793238460
INF =  float('inf')
MOD  = 1000000007
# MOD = 998244353


def bin32(num):
    return '{0:032b}'.format(num)

def add(x,y):
    return (x+y)%MOD

def sub(x,y):
    return (x-y+MOD)%MOD

def mul(x,y):
    return (x*y)%MOD

def gcd(x,y):
    if y == 0:
        return x
    return gcd(y,x%y)

def lcm(x,y):
    return (x*y)//gcd(x,y)

def power(x,y):
    res = 1
    x%=MOD
    while y!=0:
        if y&1 :
            res = mul(res,x)
        y>>=1
        x = mul(x,x)
        
    return res
        
def mod_inv(n):
    return power(n,MOD-2)

def prob(p,q):
    return mul(p,power(q,MOD-2))    
  
def ii():
    return int(input())

def li():
    return [int(i) for i in input().split()]

def ls():
    return [i for i in input().split()]

for t in range(ii()):
    t+=1
    l,r = li()
    if r%2 == 1:
        if l > r//2:
            print(r%l)
        else:
            print(r%((r+1)//2))
    else:
        x = r//2
        x+=1
        if x>=l:
            print(r%x)
        else:
            print(r%l)