#include"bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
constexpr long long mod = 1000000007;
typedef pair<int, int> P;
int res;
int fib[90];
int fi2[90][90];
int gcd(int a, int b) {
	if (a < b)swap(a, b);
	if (!b)return 0;
	if (a / b > 1&&b>1)res++;
	if (b == 1 && a > 2)res++;
	return 1+gcd(b, a%b);
}

signed main() {
	int q; cin >> q;
	fib[0]=fib[1] = 1;
	rep(i, 88)fib[i+2]=fib[i + 1] + fib[i];
	for(int i=0;i<90;i++) {
		fi2[i][0] = fi2[i][1] = 1;
		for (int j = 1; j <= 88; j++) {
			if (i == j)fi2[i][j + 1] = fi2[i][j] * 2 + fi2[i][j - 1];
			else fi2[i][j + 1] = fi2[i][j] + fi2[i][j - 1];
		}
	}

	//f[87]で10^18を超える
	rep(i, q) {
		int x, y; cin >> x >> y;
		if (x > y)swap(x, y);
		if (P(x, y) == P(1, 1)) {
			puts("1 1"); continue;
		}
		int K = 0;
		rep(i, 88) {
			if (fib[i] <= x&&fib[i + 1] <= y)K = i;
		}

		if (K == 1) {
			cout <<K<<' '<< x*y%mod << endl;
			continue;
		}

		int ans = 0;
		int d = y - fib[K + 1];
		d /= fib[K]; d++;
		ans += d;
		if (x >= fib[K + 1]) {
			d = x - fib[K + 1];
			d /= fib[K]; d++;
			ans += d;
		}

		for (int i = 1; i < K; i++) {

			if (fi2[i][K] <= x&&fi2[i][K + 1] <= y) {
				d = y - fi2[i][K + 1];
				d /= fi2[i][K]; d++;
				ans += d;
				ans %= mod;
				if (fi2[i][K + 1] <= x) {
					d = x - fi2[i][K + 1];
					d /= fi2[i][K]; d++;
					ans += d;
					ans %= mod;
				}
			}
		}

		cout <<K<<' '<< ans%mod << endl;
	}
}
### Prompt

Please create a solution in Cpp to 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"
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
constexpr long long mod = 1000000007;
typedef pair<int, int> P;
int res;
int fib[90];
int fi2[90][90];
int gcd(int a, int b) {
	if (a < b)swap(a, b);
	if (!b)return 0;
	if (a / b > 1&&b>1)res++;
	if (b == 1 && a > 2)res++;
	return 1+gcd(b, a%b);
}

signed main() {
	int q; cin >> q;
	fib[0]=fib[1] = 1;
	rep(i, 88)fib[i+2]=fib[i + 1] + fib[i];
	for(int i=0;i<90;i++) {
		fi2[i][0] = fi2[i][1] = 1;
		for (int j = 1; j <= 88; j++) {
			if (i == j)fi2[i][j + 1] = fi2[i][j] * 2 + fi2[i][j - 1];
			else fi2[i][j + 1] = fi2[i][j] + fi2[i][j - 1];
		}
	}

	//f[87]で10^18を超える
	rep(i, q) {
		int x, y; cin >> x >> y;
		if (x > y)swap(x, y);
		if (P(x, y) == P(1, 1)) {
			puts("1 1"); continue;
		}
		int K = 0;
		rep(i, 88) {
			if (fib[i] <= x&&fib[i + 1] <= y)K = i;
		}

		if (K == 1) {
			cout <<K<<' '<< x*y%mod << endl;
			continue;
		}

		int ans = 0;
		int d = y - fib[K + 1];
		d /= fib[K]; d++;
		ans += d;
		if (x >= fib[K + 1]) {
			d = x - fib[K + 1];
			d /= fib[K]; d++;
			ans += d;
		}

		for (int i = 1; i < K; i++) {

			if (fi2[i][K] <= x&&fi2[i][K + 1] <= y) {
				d = y - fi2[i][K + 1];
				d /= fi2[i][K]; d++;
				ans += d;
				ans %= mod;
				if (fi2[i][K + 1] <= x) {
					d = x - fi2[i][K + 1];
					d /= fi2[i][K]; d++;
					ans += d;
					ans %= mod;
				}
			}
		}

		cout <<K<<' '<< ans%mod << endl;
	}
}
```