#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <LL, LL> pLL;
typedef pair <int, int> pii;

inline LL Read()
{
	LL x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int n = 90;
const int mod = 1e9 + 7;

vector <pLL> a[n];
LL fib[n];

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	fib[0] = fib[1] = 1;
	for (int i = 2; i < n; i ++)
		fib[i] = fib[i - 2] + fib[i - 1];
	a[1].pb(mp(1, 2)), a[1].pb(mp(1, 3));
	for (int i = 2; i + 2 < n; i ++)
		for (auto j : a[i - 1])
			for (LL x = j.yy, y = x + j.xx; y <= fib[i + 2] + fib[i - 1]; y += j.yy)
				a[i].pb(mp(x, y));
	int Q = Read();
	while (Q --)
	{
		LL x = Read(), y = Read();
		if (x > y)
			swap(x, y);
		if (x == 1)
			printf("1 %d\n", y % mod);
		else if (y == 2)
			puts("1 4");
		else
		{
			int cur = 0, ret = 0;
			while (fib[cur + 1] <= x && fib[cur + 2] <= y)
				cur ++;
			printf("%d ", cur);
			for (auto i : a[cur - 1])
			{
				if (i.yy <= x)
					ret = ((y - i.xx) / i.yy + ret) % mod;
				if (i.yy <= y)
					ret = ((x - i.xx) / i.yy + ret) % mod;
			}
			printf("%d\n", ret);
		}
	}
}

### Prompt

Your challenge is to write 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 <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <LL, LL> pLL;
typedef pair <int, int> pii;

inline LL Read()
{
	LL x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int n = 90;
const int mod = 1e9 + 7;

vector <pLL> a[n];
LL fib[n];

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	fib[0] = fib[1] = 1;
	for (int i = 2; i < n; i ++)
		fib[i] = fib[i - 2] + fib[i - 1];
	a[1].pb(mp(1, 2)), a[1].pb(mp(1, 3));
	for (int i = 2; i + 2 < n; i ++)
		for (auto j : a[i - 1])
			for (LL x = j.yy, y = x + j.xx; y <= fib[i + 2] + fib[i - 1]; y += j.yy)
				a[i].pb(mp(x, y));
	int Q = Read();
	while (Q --)
	{
		LL x = Read(), y = Read();
		if (x > y)
			swap(x, y);
		if (x == 1)
			printf("1 %d\n", y % mod);
		else if (y == 2)
			puts("1 4");
		else
		{
			int cur = 0, ret = 0;
			while (fib[cur + 1] <= x && fib[cur + 2] <= y)
				cur ++;
			printf("%d ", cur);
			for (auto i : a[cur - 1])
			{
				if (i.yy <= x)
					ret = ((y - i.xx) / i.yy + ret) % mod;
				if (i.yy <= y)
					ret = ((x - i.xx) / i.yy + ret) % mod;
			}
			printf("%d\n", ret);
		}
	}
}

```