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
#include <cstdio>
#include <vector>
#include <cassert>
#include <algorithm>
using namespace std;

#define iter(i, n) for (int i = 1; i <= n; ++i)

const int mod = 1e9 + 7;

typedef long long i64;
typedef pair<i64, i64> pii;

#define fi first
#define se second

vector<pii> vec[100];
i64 f[100];

int F(i64 a, i64 b) {
	if (a < b) swap(a, b);
	return !b ? 0 : F(b, a % b) + 1;
}

int main() {
	f[0] = 1, f[1] = 1;
	int n = 1;

	
	while (f[n] <= 1e18) {
		++n;
		f[n] = f[n - 1] + f[n - 2];
	}
	f[n + 1] = f[n];
	--n;

	vec[0].push_back(pii(0, 1));
	vec[0].push_back(pii(0, 2));
	iter(i, n) {
		for (pii p : vec[i - 1]) {
			i64 a = p.fi, b = p.se;
		//	if (i == 4) printf("[%lld %lld] %d %d %lld\n", a, b, i-1, F(a, b), f[i+1]);
			assert(F(a, b) == i - 1);

//			printf("%d [%lld %lld]", i, a, b);
			for (; a + b <= f[i + 2]; a += b) {
				//gcd(f[i+2],f
				if (F(b, a + b) == i) vec[i].push_back(pii(b, a + b));
			}
		}
	}
	int q;
	scanf("%d", &q);
	iter(id, q) {
		i64 a, b;
		scanf("%lld%lld", &a, &b);
		if (a > b) swap(a, b);
		int ans = 1; i64 cnt = 0;
		iter(i, n) {
			if (a >= f[i] && b >= f[i + 1]) ans = i;
			else break;
		}
		printf("%d ", ans);
		for (pii p : vec[ans - 1]) {
			if (p.se <= a && p.fi + p.se <= b && p.fi != p.se) cnt = (cnt + (b - p.fi) / p.se) % mod;
			swap(a, b);
			if (p.se <= a && p.fi + p.se <= b && p.fi != p.se) cnt = (cnt + (b - p.fi) / p.se) % mod;
			swap(a, b);
			
		}
		printf("%lld\n", ans != 1 ? cnt : a * b % mod);
	}
	//
	return 0;
}
