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 CPP):

Solution:
#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
int M(int x) { return x >= mod ? x - mod : x; }
long long x, y;
long long f[102];
int res, sum, lim = 87;

bool dfs(long long a, long long b, int d) {
    if (a * f[d - 1] + b * (d > 1 ? f[d - 2] : 0) > x ||
        a * f[d] + b * f[d - 1] > y)
        return 0;
    if (d == 1) {
        if (x >= a + b) sum = M(sum + (x - b) / a % mod);
        return sum = M(sum + (y - b) / a % mod), 1;
    }

    for (int i = 1 + (!b);; ++i)
        if (!dfs(b + a * i, a, d - 1)) return i > 1;
}

int main() {
    int T;
    scanf("%d", &T), f[0] = 1, f[1] = 1;
    for (int i = 2; i <= lim; ++i) f[i] = f[i - 1] + f[i - 2];
    while (T--) {
        scanf("%lld%lld", &x, &y);
        if (x > y) swap(x, y);
        if (x == 1) {
            printf("1 %d\n", y % mod);
            continue;
        }
        for (int i = 1; i <= lim; ++i) {
            if (x < f[i - 1] || y < f[i]) {
                res = i - 2;
                break;
            }
        }
        sum = 0, dfs(1, 0, res), printf("%d %d\n", res, sum);
    }
    return 0;
}