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:
// Kenus the Ancient Greek
// * frank_c1
// * 2017 / 11 / 08

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxw = 87;
const int mo = (int)(1e9) + 7;
LL f[maxw + 10];
vector<pair<LL, LL> > g[maxw + 5];

void gen() {
	f[0] = f[1] = 1;
	for (int i = 2; i <= maxw + 3; ++i) f[i] = f[i - 1] + f[i - 2];
	g[0].push_back(make_pair(1, 1));
	for (int i = 0; i < maxw; ++i) {
		for (int j = 0; j < (int)g[i].size(); ++j) {
			LL y = g[i][j].first, x = g[i][j].second + y;
			while (x <= f[i] + f[i + 3]) {
				g[i + 1].push_back(make_pair(x, y)); x += y;
			}
		}
	}
}

LL calc(LL n, LL m, int p) {
	LL ret = 0;
	for (int j = 0; j < (int)g[p - 1].size(); ++j) {
		LL x = g[p - 1][j].first, y = g[p - 1][j].second;
		if (x <= m) (ret += (n - y) / x) %= mo;
	} return ret;
}

void solve() {
	LL n, m; 
	scanf("%lld%lld", &n, &m);
	if (n < m) swap(n, m);
	if (n == 2 && m == 2) {
		printf("1 4\n"); return;
	}
	if (m == 1) {
		printf("1 %d\n", n % mo); return;
	}
	int p = 2, res= 0;
	for (; n >= f[p + 1] && m >= f[p]; ++p); --p;
	printf("%d ", p); res = calc(n, m, p); 
	if (f[p + 1] <= m) res += calc(m, m, p);
	printf("%d\n", res % mo);
}

int main() {
	gen();
	int T; 
	scanf("%d", &T);
	while (T--) solve(); 
	return 0;
}