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 <cstdio>
#include <algorithm>
#include <vector>
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long ll;
typedef std::pair<ll, ll> mp;

struct {
	inline operator int () { int x; return scanf("%d", &x), x; }
	inline operator ll () { ll x; return scanf("%lld", &x), x; }
	template<class T> inline void operator () (T &x) { x = *this; }
	template<class T, class ...A> inline void operator () (T &x, A &...a)
	{ x = *this; this -> operator ()(a...); }
} read;

const int maxk = 100, mod = 1000000007;
ll fib[maxk], res; int mx;
std::vector<mp> excellent[maxk];

int main() {
	int k = 1;
	fib[0] = fib[1] = 1;
	ll lim = 1000000000000000000;
	while(fib[k] + fib[k - 1] <= lim) {
		++ k;
		fib[k] = fib[k - 1] + fib[k - 2];
	}

	excellent[1].push_back(mp(1, 2));
	for(int i = 2; i <= k; i ++) {
		for(mp p : excellent[i - 1])
			excellent[i].push_back(mp(p.second, p.first + p.second));
		excellent[i].push_back(mp(fib[i + 1], fib[i + 1] + fib[i - 1]));
	}

	int q = read;
	while(q --) {
		ll x = read, y = read;
		if(x > y) std::swap(x, y);

		mx = 1;
		while(mx + 2 <= k and fib[mx + 1] <= x and fib[mx + 2] <= y)
			++ mx;

		res = 0;
		for(mp p : excellent[mx]) {
			if(p.first <= x and p.second <= y)
				res += (y - p.second) / p.first + 1;
			if(p.second <= x)
				res += (x - p.second) / p.first + 1;
			res %= mod;
		}

		if(mx == 1)
			res += x;

		printf("%d %lld\n", mx, res);
	}
}