#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXM = 90;
const long long INF = 1ll << 60;
const int MOD = 1000000000 + 7;

typedef pair<long long, long long> pii;
long long fib[MAXM];
pii spe[MAXM][MAXM];

void init()
{
	fib[0] = fib[1] = 1;
	for(int i = 2; i < MAXM; i++)
		fib[i] = fib[i - 1] + fib[i - 2];
	// for(int i = 0; i < MAXM; i++)
	// 	cerr << fib[i] << endl;
	spe[1][0] = make_pair(1, 2), spe[1][1] = make_pair(1, 3);
	for(int i = 2; i + 2 < MAXM; i++)
	{
		for(int j = 0; j <= i - 1; j++)
			spe[i][j] = make_pair(spe[i - 1][j].second, spe[i - 1][j].first + spe[i - 1][j].second);
		spe[i][i] = make_pair(fib[i], fib[i + 2]);
	}
	// for(int i = 1; i < MAXM; i++)
	// 	cerr << spe[i][i].first << ' ' << spe[i][i].second << endl;
}

long long X, Y;

void work()
{
	cin >> X >> Y;
	if(X > Y) swap(X, Y);
	int k = 1;
	while(fib[k + 1] <= X && fib[k + 2] <= Y)
		k++;
	if(k == 1)
	{
		cout << k << ' ' << (X % MOD) * (Y % MOD) % MOD << endl;
		return ;
	}
	int ans = 0;
	for(int i = 0; i <= k - 1; i++)
	{
		if(X >= spe[k - 1][i].second) ans = (ans + (Y - spe[k - 1][i].first) / spe[k - 1][i].second) % MOD;
		if(Y >= spe[k - 1][i].second) ans = (ans + (X - spe[k - 1][i].first) / spe[k - 1][i].second) % MOD;
	}
	cout << k << ' ' << ans << endl;
}	

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	init();
	int T;
	cin >> T;
	while(T--)
		work();


	return 0;
}

### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXM = 90;
const long long INF = 1ll << 60;
const int MOD = 1000000000 + 7;

typedef pair<long long, long long> pii;
long long fib[MAXM];
pii spe[MAXM][MAXM];

void init()
{
	fib[0] = fib[1] = 1;
	for(int i = 2; i < MAXM; i++)
		fib[i] = fib[i - 1] + fib[i - 2];
	// for(int i = 0; i < MAXM; i++)
	// 	cerr << fib[i] << endl;
	spe[1][0] = make_pair(1, 2), spe[1][1] = make_pair(1, 3);
	for(int i = 2; i + 2 < MAXM; i++)
	{
		for(int j = 0; j <= i - 1; j++)
			spe[i][j] = make_pair(spe[i - 1][j].second, spe[i - 1][j].first + spe[i - 1][j].second);
		spe[i][i] = make_pair(fib[i], fib[i + 2]);
	}
	// for(int i = 1; i < MAXM; i++)
	// 	cerr << spe[i][i].first << ' ' << spe[i][i].second << endl;
}

long long X, Y;

void work()
{
	cin >> X >> Y;
	if(X > Y) swap(X, Y);
	int k = 1;
	while(fib[k + 1] <= X && fib[k + 2] <= Y)
		k++;
	if(k == 1)
	{
		cout << k << ' ' << (X % MOD) * (Y % MOD) % MOD << endl;
		return ;
	}
	int ans = 0;
	for(int i = 0; i <= k - 1; i++)
	{
		if(X >= spe[k - 1][i].second) ans = (ans + (Y - spe[k - 1][i].first) / spe[k - 1][i].second) % MOD;
		if(Y >= spe[k - 1][i].second) ans = (ans + (X - spe[k - 1][i].first) / spe[k - 1][i].second) % MOD;
	}
	cout << k << ' ' << ans << endl;
}	

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	init();
	int T;
	cin >> T;
	while(T--)
		work();


	return 0;
}

```