#include <cstdio>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const ll inf = 1e18 ;
const int mo = 1e9 + 7 ;
using namespace std ;

ll f[105] ;
vector <pair <ll, ll> > P[105] ;

void pre() {
	f[0] = f[1] = 1 ;
	int n = 1 ;
	rep(i, 2, 100) {
		f[i] = f[i - 1] + f[i - 2] ;
		if (f[i] <= inf) n = i ; else break ;
	}
	P[0].push_back(make_pair(1, 1)) ;
	rep(i, 0, n - 3) {
		rep(j, 0, (int) P[i].size() - 1) {
			ll x = P[i][j].first, y = P[i][j].second ;
			for ( ; ; ) {
				x += y ;
				if (x > f[i] + f[i + 3]) break ;
				P[i + 1].push_back(make_pair(y, x)) ;
			}
		}
	}
}

int main() {
	pre() ;
	int m ;
	ll x, y ;
	scanf("%d", &m) ;
	rep(cas, 1, m) {
		scanf("%lld%lld", &x, &y) ;
		if (x > y) swap(x, y) ;
		int ans = 1 ;
		rep(i, 1, 100) if (f[i] <= x && f[i + 1] <= y) ans = i ; else break ;
		printf("%d ", ans) ;
		if (ans == 1) {
			printf("%lld\n", (ll) x * y % mo) ;
			continue ;
		}
		int k = ans ;
		ans = 0 ;
		rep(i, 0, (int) P[k - 1].size() - 1) {
			ll A = P[k - 1][i].first, B = P[k - 1][i].second ;
			if (B <= x) ans += ((y - A) / B) % mo, ans %= mo ;
			if (B <= y) ans += ((x - A) / B) % mo, ans %= mo ;
		}
		printf("%d\n", (ans + mo) % mo) ;
	}
	return 0 ;
}
### Prompt

Create a solution in cpp for 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 <cstdio>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const ll inf = 1e18 ;
const int mo = 1e9 + 7 ;
using namespace std ;

ll f[105] ;
vector <pair <ll, ll> > P[105] ;

void pre() {
	f[0] = f[1] = 1 ;
	int n = 1 ;
	rep(i, 2, 100) {
		f[i] = f[i - 1] + f[i - 2] ;
		if (f[i] <= inf) n = i ; else break ;
	}
	P[0].push_back(make_pair(1, 1)) ;
	rep(i, 0, n - 3) {
		rep(j, 0, (int) P[i].size() - 1) {
			ll x = P[i][j].first, y = P[i][j].second ;
			for ( ; ; ) {
				x += y ;
				if (x > f[i] + f[i + 3]) break ;
				P[i + 1].push_back(make_pair(y, x)) ;
			}
		}
	}
}

int main() {
	pre() ;
	int m ;
	ll x, y ;
	scanf("%d", &m) ;
	rep(cas, 1, m) {
		scanf("%lld%lld", &x, &y) ;
		if (x > y) swap(x, y) ;
		int ans = 1 ;
		rep(i, 1, 100) if (f[i] <= x && f[i + 1] <= y) ans = i ; else break ;
		printf("%d ", ans) ;
		if (ans == 1) {
			printf("%lld\n", (ll) x * y % mo) ;
			continue ;
		}
		int k = ans ;
		ans = 0 ;
		rep(i, 0, (int) P[k - 1].size() - 1) {
			ll A = P[k - 1][i].first, B = P[k - 1][i].second ;
			if (B <= x) ans += ((y - A) / B) % mo, ans %= mo ;
			if (B <= y) ans += ((x - A) / B) % mo, ans %= mo ;
		}
		printf("%d\n", (ans + mo) % mo) ;
	}
	return 0 ;
}
```