#include <bits/stdc++.h>
#define ri rd<int>
#define rl rd<LL>
#define rep(i, a, b) for (int i = (a), _ = (b); i <= _; ++ i)
#define per(i, a, b) for (int i = (a), _ = (b); i >= _; -- i)
#define For(i, a, b) for (int i = (a), _ = (b); i < _; ++ i)
using namespace std;
typedef long long LL;
const LL O = 1e9 + 7;
const int maxN = 107;

template<class T> T rd() {
	bool f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = 0;
	T x = 0; for (; isdigit(c); c = getchar()) x = x * 10 + c - 48; return f ? x : -x;
}

LL _f[maxN], *f = _f+1;
LL A, B;

LL calc1(LL a, LL b) {
	if (a > A || b > B) return 0;
	return 1 + (B - b) / a;
}

LL calc2(LL a, LL b) {
	if (a > A || b > B) return 0;
	return 1 + (A - a) / b;
}

LL calc(LL a, LL b) {
	return (calc1(a, b) + calc2(b, a)) % O;
}

int main() {

	f[-1] = f[0] = 1;
	rep (i, 1, 100) f[i] = f[i-1] + f[i-2];

	rep (tcas, 1, ri()) {
		A = rl(), B = rl();
		if (A > B) swap(A, B);

		if (B == 1) {
			puts("1 1");
			continue;
		}

		int K;
		for (K = 0; f[K] <= A && f[K+1] <= B; ++K);

		if (K == 1) {
			printf("1 %lld\n", A * B % O);
			continue;
		}

		LL res = calc(f[K-1], f[K]);
		For (k, 1, K) {
			res += calc(f[K-1] + f[K-k-2] * f[k-1], f[K] + f[K-k-1] * f[k-1]);
		}
		printf("%d %lld\n", K, (res % O + O) % O);
	}

	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
#include <bits/stdc++.h>
#define ri rd<int>
#define rl rd<LL>
#define rep(i, a, b) for (int i = (a), _ = (b); i <= _; ++ i)
#define per(i, a, b) for (int i = (a), _ = (b); i >= _; -- i)
#define For(i, a, b) for (int i = (a), _ = (b); i < _; ++ i)
using namespace std;
typedef long long LL;
const LL O = 1e9 + 7;
const int maxN = 107;

template<class T> T rd() {
	bool f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = 0;
	T x = 0; for (; isdigit(c); c = getchar()) x = x * 10 + c - 48; return f ? x : -x;
}

LL _f[maxN], *f = _f+1;
LL A, B;

LL calc1(LL a, LL b) {
	if (a > A || b > B) return 0;
	return 1 + (B - b) / a;
}

LL calc2(LL a, LL b) {
	if (a > A || b > B) return 0;
	return 1 + (A - a) / b;
}

LL calc(LL a, LL b) {
	return (calc1(a, b) + calc2(b, a)) % O;
}

int main() {

	f[-1] = f[0] = 1;
	rep (i, 1, 100) f[i] = f[i-1] + f[i-2];

	rep (tcas, 1, ri()) {
		A = rl(), B = rl();
		if (A > B) swap(A, B);

		if (B == 1) {
			puts("1 1");
			continue;
		}

		int K;
		for (K = 0; f[K] <= A && f[K+1] <= B; ++K);

		if (K == 1) {
			printf("1 %lld\n", A * B % O);
			continue;
		}

		LL res = calc(f[K-1], f[K]);
		For (k, 1, K) {
			res += calc(f[K-1] + f[K-k-2] * f[k-1], f[K] + f[K-k-1] * f[k-1]);
		}
		printf("%d %lld\n", K, (res % O + O) % O);
	}

	return 0;
}

```