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<bits/stdc++.h>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
typedef std::pair<ll, ll> pr;
const int mod = 1e9 + 7;
std::set<pr> s[100];
ll f[100], x, y;
int q;
inline void dfs(ll x, ll y, int cnt) {
	if(cnt == 89) return ;
	if(s[cnt].count(pr(x, y))) return ;
	s[cnt].emplace(x, y);
	for(int i = 1;y * i + x <= f[cnt] + f[cnt + 3];++i) {
		dfs(y, y * i + x, cnt + 1);
	}
}
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	f[0] = f[1] = 1;
	for(int i = 2;i < 93;++i) f[i] = f[i - 1] + f[i - 2];
	dfs(1, 1, 0);
	cin >> q;
	rep(i, 1, q) {
		cin >> x >> y;
		if(x > y)
			std::swap(x, y);
		int cnt = 1;
		for(;f[cnt + 1] <= x && f[cnt + 2] <= y;) 
			++ cnt;
		if(cnt == 1) {
			cout << 1 << ' ' << x * y % mod << '\n';
		} else {
			int ans = 0;
			for(const pr&i : s[cnt - 1]) {
				if(i.second <= y) ans = (ans + (x - i.first) / i.second) % mod;
				if(i.second <= x) ans = (ans + (y - i.first) / i.second) % mod;
			}
			cout << cnt << ' ' << ans << '\n';
		}
	}
}