Problem: Kenus, the organizer of International Euclidean Olympiad, is seeking a pair of two integers that requires many steps to find its greatest common divisor using the Euclidean algorithm.

You are given Q queries. The i-th query is represented as a pair of two integers X_i and Y_i, and asks you the following: among all pairs of two integers (x,y) such that 1 ≤ x ≤ X_i and 1 ≤ y ≤ Y_i, find the maximum Euclidean step count (defined below), and how many pairs have the maximum step count, modulo 10^9+7.

Process all the queries. Here, the Euclidean step count of a pair of two non-negative integers (a,b) is defined as follows:

* (a,b) and (b,a) have the same Euclidean step count.
* (0,a) has a Euclidean step count of 0.
* If a > 0 and a ≤ b, let p and q be a unique pair of integers such that b=pa+q and 0 ≤ q < a. Then, the Euclidean step count of (a,b) is the Euclidean step count of (q,a) plus 1.

Constraints

* 1 ≤ Q ≤ 3 × 10^5
* 1 ≤ X_i,Y_i ≤ 10^{18}(1 ≤ i ≤ Q)

Input

The input is given from Standard Input in the following format:


Q
X_1 Y_1
:
X_Q Y_Q


Output

For each query, print the maximum Euclidean step count, and the number of the pairs that have the maximum step count, modulo 10^9+7, with a space in between.

Examples

Input

3
4 4
6 10
12 11


Output

2 4
4 1
4 7


Input

10
1 1
2 2
5 1000000000000000000
7 3
1 334334334334334334
23847657 23458792534
111111111 111111111
7 7
4 19
9 10


Output

1 1
1 4
4 600000013
3 1
1 993994017
35 37447
38 2
3 6
3 9
4 2

Solution (in PYTHON3):

Solution:
from math import gcd
import random,time

def gcdcount(n,m):
    x,y=min(n,m),max(n,m)
    if x==0:
        return 0
    else:
        return 1+gcdcount(x,y%x)

fibo=[0,1,2]
for i in range(100):
    fibo.append(fibo[-1]+fibo[-2])

gcdfibo=[[],[(1,2)],[(2,3)],[(3,5),(4,7)]]
for i in range(4,101):
    temp=[]
    for a,b in gcdfibo[-1]:
        temp.append((b,a+b))
    b=gcdfibo[-1][1][1]
    a=fibo[i]
    temp.append((b,a+b))
    temp.sort()
    gcdfibo.append(temp)

def solve(x,y):
    if x>y:
        x,y=y,x
    mod=10**9+7
    if x==1:
        return  (1,y%mod)
    if (x,y)==(2,2):
        return (1,3)
    id=1
    t=-1
    while True:
        if fibo[id+1]<=x and fibo[id+2]<=y:
            id+=1
        else:
            if fibo[id+1]>x:
                if fibo[id+2]>y:
                    t=0
                else:
                    t=1
            else:
                t=2
            break

    #print(t)
    if t==0:
        res=0
        for a,b in gcdfibo[id]:
            if a<=x and b<=y:
                res+=1
        return (id,res)
    elif t==1:
        res=0
        for a,b in gcdfibo[id]:
            if a<=x:
                res+=(y-b)//a+1
                res%=mod
        return (id,res)
    else:
        res=0
        for a,b in gcdfibo[id+1]:
            u,v=b-a,a
            if u<=x and v<=y:
                res+=1
                res%=mod
            if u<=y and v<=x:
                res+=1
                res%=mod
        return (id,res)

mod=10**9+7
Q=int(input())
start=time.time()
for _ in range(Q):
    x,y=map(int,input().split())
    id,res=solve(x,y)
    if (x,y)==(2,2):
        res+=1
    print(id,res)

#print(time.time()-start)