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:
#include <bits/stdc++.h>
#define REP(i, x, y) for(int i = (int)x; i <= (int)y; i ++)
#define FOR(i, x, y) for(int i = (int)x; i <  (int)y; i ++)
#define PER(i, x, y) for(int i = (int)x; i >= (int)y; i --)
#define fi first
#define se second
#define mk make_pair
#define pb push_back
using namespace std;
typedef long long LL;
const	int P = 1e9 + 7;

LL Fib[10005];
int dp[55][55], mx[55][55];
pair<LL, LL> bas[105][105];

void work(LL A, LL B){
	if(min(A, B) == 1){
		printf("1 %lld\n", max(A, B) % P);
		return;
	}
	if(A == 2 && B == 2){
		printf("1 4\n");
		return;
	}
	int ans = 2;
	LL res = 0;
	for(ans = 2; Fib[ans] <= B && Fib[ans + 1] <= A; ans ++);
	ans --;
	printf("%d ", ans);
	REP(k, 1, ans){
		LL u = bas[ans][k].fi, v = bas[ans][k].se;
		//du <= Xi
		//dv + dku <= Yi
		LL C = A, D = B;
		for(LL d = 1; d * u <= C; d ++){
			LL lim = (D - d * v);
			if(lim < 0) break;
			res += (lim / (d * u) + 1) % P;
			res %= P;
		}
		swap(C, D);
		for(LL d = 1; d * u <= C; d ++){
			LL lim = (D - d * v);
			if(lim < 0) break;
			res += (lim / (d * u) + 1) % P;
			res %= P;
		}
	}
	printf("%lld\n", res);
}

int main(){
	Fib[1] = 1;
	Fib[0] = 1;
	REP(i, 2, 87) Fib[i] = Fib[i - 1] + Fib[i - 2];
	bas[2][1] = mk(2LL, 3LL);
	bas[2][2] = mk(3LL, 4LL);
	REP(i, 3, 86){
		FOR(j, 1, i){
			LL u = bas[i - 1][j].fi, v = bas[i - 1][j].se;
			bas[i][j] = mk(v, v + u);
			if(j == 1) bas[i][i] = mk(v + u, v + u + u);
		}
	}
	int Q;
	scanf("%d", &Q);
	while(Q --){
		LL X, Y;
		scanf("%lld%lld", &X, &Y);
		if(X < Y) swap(X, Y);
		work(X, Y);
	}
	return 0;
}