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 EB emplace_back

typedef long long ll;
typedef std::pair <ll, ll> pr;
typedef std::vector <pr> vector;
const int mod = 1000000007;

int q;
ll A, B, f[89];
vector G[89];

int main() {
	int i; ll ans = 0;
	for (f[1] = f[2] = 1, i = 3; i < 89; ++i) f[i] = f[i - 1] + f[i - 2];
	G[2].EB(2, 3), G[2].EB(3, 4);
	for (i = 3; i < 87; ++i) {
		G[i].reserve(i);
		for (const pr &p : G[i - 1]) G[i].EB(p.second, p.first + p.second);
		G[i].EB(f[i + 2], f[i + 2] + f[i]);
	}
	for (scanf("%d", &q); q; --q) {
		scanf("%lld%lld", &A, &B), ans = 0;
		if (A > B) std::swap(A, B);
		if (A == 1 || B == 2) {printf("1 %d\n", int(A * B % mod)); continue;}
		for (i = 2; f[i + 1] <= A && f[i + 2] <= B; ++i); --i;
		for (const pr &p : G[i]) {
			if (p.first <= A && p.second <= B) ans += ((B - p.second) / p.first + 1) % mod;
			if (p.first <= B && p.second <= A) ans += ((A - p.second) / p.first + 1) % mod;
		}
		printf("%d %d\n", i, int(ans % mod));
	}
	return 0;
}