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 <algorithm>
#include <utility>
#include <cstdio>
#include <vector>
#define pb push_back
#define MOD 1000000007
typedef long long ll;
ll F[105];
std::vector<std::pair<ll, ll> > vec[105];
int main()
{
	// freopen("AGC015-F.in", "r", stdin);
	int len = 2;
	F[0] = F[1] = 1;
	while (true)
	{
		F[len] = F[len - 1] + F[len - 2];
		if (F[len] > 2e18)
			break;
		len++;
	}
	vec[1].pb({1, 2});
	vec[1].pb({1, 3});
	for (int i = 1; i < len - 3; i++)
	{
		for (auto &it : vec[i])
		{
			auto x = it.first, y = it.second;
			while (x + y <= F[i + 3])
			{
				x += y;
				vec[i + 1].pb({y, x});
			}
		}
	}
	int cas;
	scanf("%d", &cas);
	while (cas--)
	{
		ll n, m;
		scanf("%lld%lld", &n, &m);
		if (n > m)
			std::swap(n, m);
		int ans = 1;
		while (ans + 2 < len && F[ans + 1] <= n && F[ans + 2] <= m)
			ans++;
		if (ans == 1)
			printf("1 %lld\n", n * m % MOD);
		else
		{
			ll cnt = 0;
			for (auto &it : vec[ans - 1])
			{
				auto x = it.first, y = it.second;
				if (y <= n)
					(cnt += (m - x) / y) %= MOD;
				if (y <= m)
					(cnt += (n - x) / y) %= MOD;
			}
			printf("%d %lld\n", ans, cnt);
		}
	}
	return 0;
}