#include <bits/stdc++.h>

typedef long long LL;
typedef std::pair<LL, LL> pi;
const int mod = 1000000007;
void reduce(int & x) { x += x >> 31 & mod; }
int remod(LL x) { x %= mod; return x + (x >> 63 & mod); }
const int MAXN = 100;
LL fib[MAXN]; int bak;
pi li[MAXN][MAXN];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	fib[1] = 1;
	for (bak = 1; fib[bak] <= 1e18; )
		fib[bak + 1] = fib[bak] + fib[bak - 1], ++bak;
	li[1][1] = pi(1, 2);
	for (int i = 2; i < bak; ++i) {
		for (int j = 1; j < i; ++j) {
			pi t = li[i - 1][j];
			li[i][j] = pi(t.second, t.first + t.second);
		}
		li[i][i] = pi(fib[i + 2], fib[i] + fib[i + 2]);
	}
	int Q; std::cin >> Q;
	while (Q --> 0) {
		LL x, y; std::cin >> x >> y;
		if (x > y) std::swap(x, y);
		int ax = std::upper_bound(fib + 2, fib + bak + 1, x) - fib - 2;
		int ay = std::upper_bound(fib + 2, fib + bak + 1, y) - fib - 2;
		int K = std::max(std::min(ax, ay - 1), 1);
		int ans = 0;
		for (int i = 1; i <= K; ++i) {
			pi t = li[K][i];
			for (int j = 0; j != 2; ++j) {
				if (t.first <= x && t.second <= y)
					reduce(ans += remod((y - t.second) / t.first + 1) - mod);
				std::swap(t.first, t.second);
			}
		}
		if (K == 1) reduce(ans += remod(x) - mod);
		std::cout << K << ' ' << ans << '\n';
	}
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>

typedef long long LL;
typedef std::pair<LL, LL> pi;
const int mod = 1000000007;
void reduce(int & x) { x += x >> 31 & mod; }
int remod(LL x) { x %= mod; return x + (x >> 63 & mod); }
const int MAXN = 100;
LL fib[MAXN]; int bak;
pi li[MAXN][MAXN];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	fib[1] = 1;
	for (bak = 1; fib[bak] <= 1e18; )
		fib[bak + 1] = fib[bak] + fib[bak - 1], ++bak;
	li[1][1] = pi(1, 2);
	for (int i = 2; i < bak; ++i) {
		for (int j = 1; j < i; ++j) {
			pi t = li[i - 1][j];
			li[i][j] = pi(t.second, t.first + t.second);
		}
		li[i][i] = pi(fib[i + 2], fib[i] + fib[i + 2]);
	}
	int Q; std::cin >> Q;
	while (Q --> 0) {
		LL x, y; std::cin >> x >> y;
		if (x > y) std::swap(x, y);
		int ax = std::upper_bound(fib + 2, fib + bak + 1, x) - fib - 2;
		int ay = std::upper_bound(fib + 2, fib + bak + 1, y) - fib - 2;
		int K = std::max(std::min(ax, ay - 1), 1);
		int ans = 0;
		for (int i = 1; i <= K; ++i) {
			pi t = li[K][i];
			for (int j = 0; j != 2; ++j) {
				if (t.first <= x && t.second <= y)
					reduce(ans += remod((y - t.second) / t.first + 1) - mod);
				std::swap(t.first, t.second);
			}
		}
		if (K == 1) reduce(ans += remod(x) - mod);
		std::cout << K << ' ' << ans << '\n';
	}
	return 0;
}

```