#include <cstdio>
#include <vector>
#include <algorithm>
const long long MOD = 1000000007;
long long fib[88];
std::vector < std::pair < long long, long long > > V[87];
bool Under(long long a, long long b, long long x, long long y, int d)
{
	if (d == 0)
		return a <= x && b <= y;
	if (d == 1)
		return b <= x && a + b <= y;
	return fib[d - 2] * a + fib[d - 1] * b <= x && fib[d - 1] * a + fib[d] * b <= y;
}
void List(std::vector < std::pair < long long, long long > > &V, long long a, long long b, long long x, long long y, int d)
{
	if (d == 1)
	{
		V.push_back({ a, b });
		return;
	}
	for (long long i = a + b; ; i += b)
		if (Under(b, i, x, y, d - 1))
			List(V, b, i, x, y, d - 1);
		else
			return;
}
long long count(long long x, long long y, int d)
{
	long long O = 0;
	for (auto i : V[d])
		if (i.second <= x && i.first + i.second <= y)
			O = (O + (y - i.first) / i.second + std::max(x - i.first, i.second - 1) / i.second) % MOD;
	return O;
}
int main()
{
	fib[0] = 1;
	fib[1] = 1;
	fib[2] = 2;
	fib[3] = 3;
	for (int i = 4; i <= 87; i++)
		fib[i] = fib[i - 1] + fib[i - 2];
	for (int i = 2; i <= 85; i++)
	{
		long long x = fib[i + 2] - 1, y = fib[i + 2] - 1;
		for (long long j = 2; ; j++)
			if (Under(1, j, x, y, i - 1))
				List(V[i], 1, j, x, y, i - 1);
			else
				break;
	}
	int Q;
	long long x, y;
	for (scanf("%d", &Q); Q--; )
	{
		scanf("%lld%lld", &x, &y);
		if (x > y)
			std::swap(x, y);
		for (int i = 85; ; i--)
		{
			if (i == 1)
			{
				printf("1 %lld\n", x * y % MOD);
				break;
			}
			if (x >= fib[i] && y >= fib[i + 1])
			{
				printf("%d %lld\n", i, count(x, y, i));
				break;
			}
		}
	}
	return 0;
}

### Prompt

In Cpp, your task is to solve 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 <cstdio>
#include <vector>
#include <algorithm>
const long long MOD = 1000000007;
long long fib[88];
std::vector < std::pair < long long, long long > > V[87];
bool Under(long long a, long long b, long long x, long long y, int d)
{
	if (d == 0)
		return a <= x && b <= y;
	if (d == 1)
		return b <= x && a + b <= y;
	return fib[d - 2] * a + fib[d - 1] * b <= x && fib[d - 1] * a + fib[d] * b <= y;
}
void List(std::vector < std::pair < long long, long long > > &V, long long a, long long b, long long x, long long y, int d)
{
	if (d == 1)
	{
		V.push_back({ a, b });
		return;
	}
	for (long long i = a + b; ; i += b)
		if (Under(b, i, x, y, d - 1))
			List(V, b, i, x, y, d - 1);
		else
			return;
}
long long count(long long x, long long y, int d)
{
	long long O = 0;
	for (auto i : V[d])
		if (i.second <= x && i.first + i.second <= y)
			O = (O + (y - i.first) / i.second + std::max(x - i.first, i.second - 1) / i.second) % MOD;
	return O;
}
int main()
{
	fib[0] = 1;
	fib[1] = 1;
	fib[2] = 2;
	fib[3] = 3;
	for (int i = 4; i <= 87; i++)
		fib[i] = fib[i - 1] + fib[i - 2];
	for (int i = 2; i <= 85; i++)
	{
		long long x = fib[i + 2] - 1, y = fib[i + 2] - 1;
		for (long long j = 2; ; j++)
			if (Under(1, j, x, y, i - 1))
				List(V[i], 1, j, x, y, i - 1);
			else
				break;
	}
	int Q;
	long long x, y;
	for (scanf("%d", &Q); Q--; )
	{
		scanf("%lld%lld", &x, &y);
		if (x > y)
			std::swap(x, y);
		for (int i = 85; ; i--)
		{
			if (i == 1)
			{
				printf("1 %lld\n", x * y % MOD);
				break;
			}
			if (x >= fib[i] && y >= fib[i + 1])
			{
				printf("%d %lld\n", i, count(x, y, i));
				break;
			}
		}
	}
	return 0;
}

```