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 debug(...) fprintf(stderr, __VA_ARGS__)
#define int long long
#define MP make_pair
#define N 805

using namespace std;
typedef pair<int, int> Pair;

inline int input() {
	int x; char ch; while (!isdigit(ch = getchar()));
	for (x = ch ^ 48; isdigit(ch = getchar()); x = x * 10 + (ch ^ 48));
	return x;
}

const int mod = 1e9 + 7;
const int MAX = 88;
int Q, x, y;
int f[100];

vector<Pair> P[100]; // P[i] 存的是 f(x, y) == i 的 优秀的对

void init() {
	f[0] = 1, f[1] = 2;
	for (int i = 2; i <= MAX; ++i) f[i] = f[i - 1] + f[i - 2];
	P[1].push_back(MP(1, 2)), P[1].push_back(MP(1, 3)), P[1].push_back(MP(1, 4));

	for (int i = 1; i <= MAX; ++i)
		for (auto p : P[i]) {
			int x = p.first, y = p.second;
			for (int j = x + y; j <= f[i + 1] << 1; j += y)
				P[i + 1].push_back(MP(y, j));
		}
}

void solve(int x, int y) {
	int k = 1;
	if (x > y) swap(x, y);
	while (f[k] <= x && f[k + 1] <= y) ++k;
	printf("%lld ", k);

	if (k == 1) {
		printf("%lld\n", x % mod * y % mod);
		return void();
	}

	int res = 0;
	for (auto p : P[k - 1]) {
		int xd = p.first, yd = p.second;
		if (yd <= x) res += (y - xd) / yd % mod;
		if (yd <= y) res += (x - xd) / yd % mod;
	}

	printf("%lld\n", res % mod);
}

signed main() {

	init();
	Q = input();
	while (Q--) {
		x = input(), y = input();
		solve(x, y);
	}
	return 0;
}