#include <bits/stdc++.h>
using namespace std;

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

int qn, tot;
long long n, m, fib[100];
vector<pair<long long, long long> > a[100];

int main() {
	fib[0] = fib[1] = 1;
	tot = 1;
	while (1) {
		fib[tot + 1] = fib[tot] + fib[tot - 1];
		if (fib[tot + 1] > inf) break;
		++tot;
	}
	a[1].push_back(make_pair(1, 2));
	a[1].push_back(make_pair(1, 3));
	a[1].push_back(make_pair(1, 4));
	for (int i = 1; i < tot - 2; ++i) {
		vector<pair<long long, long long> >::iterator it;
		for (it = a[i].begin(); it != a[i].end(); ++it) {
			long long x = it->first, y = it->second;
			x += y;
			while (x <= fib[i + 3] + fib[i]) {
				a[i + 1].push_back(make_pair(y, x));
				x += y;
			}
		}
	}
	for (scanf("%d", &qn); qn; --qn) {
		scanf("%lld%lld", &n, &m);
		if (n > m) swap(n, m);
		int ans_max = 1;
		while (fib[ans_max + 1] <= n && fib[ans_max + 2] <= m) ++ans_max;
		printf("%d ", ans_max);
		if (ans_max == 1)
			printf("%d\n", static_cast<int>(n * m % mod));
		else {
			long long cnt = 0;
			vector<pair<long long, long long> >::iterator it;
			for (it = a[ans_max - 1].begin(); it != a[ans_max - 1].end(); ++it) {
				long long x = it->first, y = it->second;
				if (y <= n) (cnt += (m - x) / y) %= mod;
				if (y <= m) (cnt += (n - x) / y) %= mod;
			}
			printf("%d\n", static_cast<int>(cnt % mod));
		}
	}
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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;

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

int qn, tot;
long long n, m, fib[100];
vector<pair<long long, long long> > a[100];

int main() {
	fib[0] = fib[1] = 1;
	tot = 1;
	while (1) {
		fib[tot + 1] = fib[tot] + fib[tot - 1];
		if (fib[tot + 1] > inf) break;
		++tot;
	}
	a[1].push_back(make_pair(1, 2));
	a[1].push_back(make_pair(1, 3));
	a[1].push_back(make_pair(1, 4));
	for (int i = 1; i < tot - 2; ++i) {
		vector<pair<long long, long long> >::iterator it;
		for (it = a[i].begin(); it != a[i].end(); ++it) {
			long long x = it->first, y = it->second;
			x += y;
			while (x <= fib[i + 3] + fib[i]) {
				a[i + 1].push_back(make_pair(y, x));
				x += y;
			}
		}
	}
	for (scanf("%d", &qn); qn; --qn) {
		scanf("%lld%lld", &n, &m);
		if (n > m) swap(n, m);
		int ans_max = 1;
		while (fib[ans_max + 1] <= n && fib[ans_max + 2] <= m) ++ans_max;
		printf("%d ", ans_max);
		if (ans_max == 1)
			printf("%d\n", static_cast<int>(n * m % mod));
		else {
			long long cnt = 0;
			vector<pair<long long, long long> >::iterator it;
			for (it = a[ans_max - 1].begin(); it != a[ans_max - 1].end(); ++it) {
				long long x = it->first, y = it->second;
				if (y <= n) (cnt += (m - x) / y) %= mod;
				if (y <= m) (cnt += (n - x) / y) %= mod;
			}
			printf("%d\n", static_cast<int>(cnt % mod));
		}
	}
}
```