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

Solution (in CPP):

Solution:
//Created Time:2019/12/9 8:57:49
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define mod 1000000007
#define ll long long

using namespace std;

ll f[100];

void init();
void work(ll, ll);
int calc(ll, ll, ll, ll);
ll read();

int main(){
	init(); int t; cin >> t;
	while(t--){
		ll x = read(), y = read();
		if(x > y) swap(x, y); work(x, y);
	}
	return 0;
}

void init(){
	f[1] = 1;
	for(int i = 2; i <= 88; ++i)
		f[i] = f[i - 1] + f[i - 2];
	return ;
}

int calc(ll x, ll y, ll a, ll b){
	int ans = ((y - b) / a + 1) % mod;
	if(b <= x) (ans += ((y - a) / b + 1) % mod) %= mod;
	return ans;
}

void work(ll x, ll y){
	if(x == 1){
		printf("%d %d\n", 1, y % mod);
		return ;
	}
	if(x == 2 && y == 2){
		puts("1 4");
		return ;
	}
	int pos = upper_bound(f + 1, f + 89, x) - f - 1;
	if(f[pos + 1] > y) --pos; printf("%d ", pos - 1);
	ll a = f[pos], b = f[pos + 1];
	int ans = calc(x, y, a, b);
	for(int i = 2; i < pos; ++i){
		ll na = a + f[pos - i] * f[i], nb = b + f[pos - i + 1] * f[i];
		if(na < a || nb < b || na > x || nb > y) continue;
		(ans += calc(x, y, na, nb)) %= mod;
	}
	printf("%d\n", ans);
	return ;
}

ll read(){
	ll x = 0; char ch = getchar();
	for(; !isdigit(ch); ch = getchar());
	for(; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
	return x;
}