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:
q = int(input())
m = 10**9+7
for _ in range(q):
    x, y = sorted(map(int, input().split()))
    s = 0
    i = j = 1
    while j <= x and i+j <= y:
        i, j = j, i+j
        s += 1
    if x == 1 or y <= 2:
        print(1, x*y % m)
    else:
        a = 0
        a += max(0, (y-j) // i+1)
        a += max(0, (x-j) // i+1)
        i = j = 1
        for c in range(s-1):
            k, l = j, i+j*2
            for _ in range(s-c-1):
                k, l = l, k+l
            if x >= k:
                a += max(0, (y-l) // k+1)
            if y >= k:
                a += max(0, (x-l) // k+1)
            i, j = j, i+j
        print(s, a % m)