/*
	最后一步一定是 (1, 0)
	从小往大搜, 多加剪枝
*/
#include <iostream>
#include <cstdio>

using namespace std;

typedef long long LL;

const LL inf = 1e18;
const int mod = 1e9 + 7;

int res;
LL ans, x, y;
LL fib[100] = {0, 1};

LL tA, tB;
bool dfs(int dep, LL A, LL B) {
	if(A * fib[dep + 1] + B * fib[dep] > y) return false;
	if(A * fib[dep] + B * fib[dep - 1] > x) return false;
	if(dep == 1) {
		ans = (ans + (y - B) / A) % mod;
		if(A <= y && B <= x)
			ans = (ans + (x - B) / A) % mod;
		return true;
	}
	for(int i = 1 + (!B); true; ++i) {
		tB = A, tA = i * A + B;
		if(!dfs(dep - 1, tA, tB))
			return i > 1;
	}
}

int main() {
//	freopen("fibo.in", "r", stdin);
//	freopen("fibo.out", "w", stdout);
	for(int i = 1; fib[i] < inf; ++i)
		fib[i + 1] = fib[i] + fib[i - 1];
	
	int q;
	scanf("%d", &q);
	for(int tim = 1; tim <= q; ++tim) {
		scanf("%lld%lld", &x, &y);
		if(x > y) swap(x, y); // x <= y

		int o = 0;
		while(fib[o + 1] <= x) ++o;
		if(fib[o + 1] > y) --o;
		
		ans = 0;
		res = o - 1;
		if(!res) res = 1;
		
		if(res == 1) {
			printf("1 %lld\n", x == 2 ? 4 : y % mod);
			continue;
		}
		
		dfs(res, 1, 0);
		printf("%d %lld\n", res, ans);
	}
	return 0;
}
### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
/*
	最后一步一定是 (1, 0)
	从小往大搜, 多加剪枝
*/
#include <iostream>
#include <cstdio>

using namespace std;

typedef long long LL;

const LL inf = 1e18;
const int mod = 1e9 + 7;

int res;
LL ans, x, y;
LL fib[100] = {0, 1};

LL tA, tB;
bool dfs(int dep, LL A, LL B) {
	if(A * fib[dep + 1] + B * fib[dep] > y) return false;
	if(A * fib[dep] + B * fib[dep - 1] > x) return false;
	if(dep == 1) {
		ans = (ans + (y - B) / A) % mod;
		if(A <= y && B <= x)
			ans = (ans + (x - B) / A) % mod;
		return true;
	}
	for(int i = 1 + (!B); true; ++i) {
		tB = A, tA = i * A + B;
		if(!dfs(dep - 1, tA, tB))
			return i > 1;
	}
}

int main() {
//	freopen("fibo.in", "r", stdin);
//	freopen("fibo.out", "w", stdout);
	for(int i = 1; fib[i] < inf; ++i)
		fib[i + 1] = fib[i] + fib[i - 1];
	
	int q;
	scanf("%d", &q);
	for(int tim = 1; tim <= q; ++tim) {
		scanf("%lld%lld", &x, &y);
		if(x > y) swap(x, y); // x <= y

		int o = 0;
		while(fib[o + 1] <= x) ++o;
		if(fib[o + 1] > y) --o;
		
		ans = 0;
		res = o - 1;
		if(!res) res = 1;
		
		if(res == 1) {
			printf("1 %lld\n", x == 2 ? 4 : y % mod);
			continue;
		}
		
		dfs(res, 1, 0);
		printf("%d %lld\n", res, ans);
	}
	return 0;
}
```