#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll fib[100];
typedef pair<ll, ll>pii;
vector<pii>dat[100];
ll mod = 1000000007;
void calc(ll a, ll b, ll k)
{
	if (b > 1000000000000000000LL)return;
	if (a > fib[k + 2] || b > fib[k + 2])return;
	dat[k].push_back(make_pair(a, b));
	dat[k].push_back(make_pair(b, a));
	calc(b, a + b, k + 1);
	calc(a + b + b, a + b + b + b, k + 2);
}
int main()
{
	/*for (int i = 1; i <= 34; i++)
	{
		for (int j = 1; j <= 34; j++)
		{
			int t = 1;
			int x = i, y = j;
			for (;;)
			{
				if (x < y)swap(x, y);
				x %= y;
				if (x == 0)break;
				t++;
			}
			printf("%d ", t);
		}
		printf("\n");
	}*/
	fib[1] = 1;
	for (int i = 2; i <= 90; i++)fib[i] = fib[i - 1] + fib[i - 2];
	calc(1, 1, 1);
	int query;
	scanf("%d", &query);
	for (int p = 0; p < query; p++)
	{
		ll za, zb;
		scanf("%lld%lld", &za, &zb);
		if (za > zb)swap(za, zb);
		if (za == 1)
		{
			printf("1 %lld\n", zb % mod);
			continue;
		}
		if (za == 2 && zb == 2)
		{
			printf("1 4\n");
			continue;
		}
		int ans = 0;
		for (int t = 0;; t++)
		{
			if (za >= fib[t] && zb >= fib[t + 1])ans = t;
			else break;
		}
		ll cnt = 0;
		for (int i = 0; i < dat[ans].size(); i++)
		{
			if (dat[ans][i].first <= za)cnt += max(0LL, (zb - dat[ans][i].second + dat[ans][i].first) / dat[ans][i].first);
			cnt %= mod;
		}
		printf("%d %lld\n", ans - 1, cnt);
	}
}
### 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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll fib[100];
typedef pair<ll, ll>pii;
vector<pii>dat[100];
ll mod = 1000000007;
void calc(ll a, ll b, ll k)
{
	if (b > 1000000000000000000LL)return;
	if (a > fib[k + 2] || b > fib[k + 2])return;
	dat[k].push_back(make_pair(a, b));
	dat[k].push_back(make_pair(b, a));
	calc(b, a + b, k + 1);
	calc(a + b + b, a + b + b + b, k + 2);
}
int main()
{
	/*for (int i = 1; i <= 34; i++)
	{
		for (int j = 1; j <= 34; j++)
		{
			int t = 1;
			int x = i, y = j;
			for (;;)
			{
				if (x < y)swap(x, y);
				x %= y;
				if (x == 0)break;
				t++;
			}
			printf("%d ", t);
		}
		printf("\n");
	}*/
	fib[1] = 1;
	for (int i = 2; i <= 90; i++)fib[i] = fib[i - 1] + fib[i - 2];
	calc(1, 1, 1);
	int query;
	scanf("%d", &query);
	for (int p = 0; p < query; p++)
	{
		ll za, zb;
		scanf("%lld%lld", &za, &zb);
		if (za > zb)swap(za, zb);
		if (za == 1)
		{
			printf("1 %lld\n", zb % mod);
			continue;
		}
		if (za == 2 && zb == 2)
		{
			printf("1 4\n");
			continue;
		}
		int ans = 0;
		for (int t = 0;; t++)
		{
			if (za >= fib[t] && zb >= fib[t + 1])ans = t;
			else break;
		}
		ll cnt = 0;
		for (int i = 0; i < dat[ans].size(); i++)
		{
			if (dat[ans][i].first <= za)cnt += max(0LL, (zb - dat[ans][i].second + dat[ans][i].first) / dat[ans][i].first);
			cnt %= mod;
		}
		printf("%d %lld\n", ans - 1, cnt);
	}
}
```