#include "bits/stdc++.h"
using namespace std;

constexpr long long INF = 1000000000000000000, MOD = 1000000007;

int main() {
	vector<long long> Fib;
	Fib.push_back(1), Fib.push_back(1);
	while (1) {
		Fib.push_back(Fib[Fib.size() - 2] + Fib[Fib.size() - 1]);
		if (Fib[Fib.size() - 1] > INF) {
			Fib.pop_back();
			break;
		}
	}
	vector<vector<pair<long long, long long> > > Pair(Fib.size());
	Pair[1].push_back({ 1, 2 });
	for (int i = 1; i < Fib.size() - 1; i++) {
		for (int j = 0; j < Pair[i].size(); j++) {
			Pair[i + 1].push_back({ Pair[i][j].second, Pair[i][j].first + Pair[i][j].second });
		}
		Pair[i + 1].push_back({ Pair[i][0].first + Pair[i][0].second, Pair[i][0].first + Pair[i][0].first + Pair[i][0].second });
	}
	int Q;
	cin >> Q;
	for (int q = 0; q < Q; q++) {
		long long X, Y;
		cin >> X >> Y;
		if (X > Y) swap(X, Y);
		if (X == 1 || Y == 2) {
			cout << 1 << ' ' << (X * Y) % MOD << endl;
			continue;
		}
		int A = -1;
		for (int i = 0; ; i++) {
			if (Fib[i] <= X && Fib[i + 1] <= Y) A++;
			else break;
		}
		cout << A << ' ';
		long long B = 0;
		for (int i = 0; i < A; i++) {
			long long P = Pair[A][i].first, Q = Pair[A][i].second;
			if (P <= X) B += (Y - Q + P) / P, B %= MOD;
			if (Q <= X) B += (X - Q + P) / P, B %= MOD;
		}
		cout << B << endl;
	}
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#include "bits/stdc++.h"
using namespace std;

constexpr long long INF = 1000000000000000000, MOD = 1000000007;

int main() {
	vector<long long> Fib;
	Fib.push_back(1), Fib.push_back(1);
	while (1) {
		Fib.push_back(Fib[Fib.size() - 2] + Fib[Fib.size() - 1]);
		if (Fib[Fib.size() - 1] > INF) {
			Fib.pop_back();
			break;
		}
	}
	vector<vector<pair<long long, long long> > > Pair(Fib.size());
	Pair[1].push_back({ 1, 2 });
	for (int i = 1; i < Fib.size() - 1; i++) {
		for (int j = 0; j < Pair[i].size(); j++) {
			Pair[i + 1].push_back({ Pair[i][j].second, Pair[i][j].first + Pair[i][j].second });
		}
		Pair[i + 1].push_back({ Pair[i][0].first + Pair[i][0].second, Pair[i][0].first + Pair[i][0].first + Pair[i][0].second });
	}
	int Q;
	cin >> Q;
	for (int q = 0; q < Q; q++) {
		long long X, Y;
		cin >> X >> Y;
		if (X > Y) swap(X, Y);
		if (X == 1 || Y == 2) {
			cout << 1 << ' ' << (X * Y) % MOD << endl;
			continue;
		}
		int A = -1;
		for (int i = 0; ; i++) {
			if (Fib[i] <= X && Fib[i + 1] <= Y) A++;
			else break;
		}
		cout << A << ' ';
		long long B = 0;
		for (int i = 0; i < A; i++) {
			long long P = Pair[A][i].first, Q = Pair[A][i].second;
			if (P <= X) B += (Y - Q + P) / P, B %= MOD;
			if (Q <= X) B += (X - Q + P) / P, B %= MOD;
		}
		cout << B << endl;
	}
}

```