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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)
#define x first
#define y second
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<LL, LL> PII;

const LL MAX = 1e18 + 10;
const int N = 100 + 5;

int n;
LL f[N];
vector<PII> P[N];

void init(){
	f[0] = f[1] = 1;
	n = 1;
	while(f[n] < MAX) ++n, f[n] = f[n - 1] + f[n - 2];
	f[n + 1] = f[n] + f[n - 1];
	--n;
	P[0].pb(mp(0, 1));
	P[1].pb(mp(1, 2)), P[1].pb(mp(1, 3));

	For(i, 1, n - 1){
		for(PII s : P[i]){
			LL u = s.x, v = s.y;
			//printf("(%lld, %lld) ", u, v);
			if(u == v) continue;
			if(u > v) swap(u, v);
			u += v;
			while(u <= f[i + 3]) P[i + 1].pb(mp(u, v)), u += v;
		}
		//puts("");
	}

	For(i, 1, n){
		sort(P[i].begin(), P[i].end());
		P[i].resize(unique(P[i].begin(), P[i].end()) - P[i].begin());
	}
}

const int Mod = 1e9 + 7;

int main(){
	init();
	int q;
	scanf("%d", &q);
	while(q--){
		LL x, y;
		scanf("%lld%lld", &x, &y);
		if(x > y) swap(x, y);

		int ans = 0;
		while(f[ans + 1] <= x && f[ans + 2] <= y) ++ans;

		if(ans <= 1){
			printf("1 %lld\n", x % Mod * (y % Mod) % Mod);
			continue;
		}
		LL cnt = 0;

		for(PII p : P[ans]){
			LL u = p.x, v = p.y;
			if(u < v) swap(u, v);
			if(u / v >= 2) continue;
			//printf("%lld %lld\n", u, v);
			if(v <= y && x >= u) (cnt += (x - u + v) / v) %= Mod;
			if(v <= x && y >= u) (cnt += (y - u + v) / v) %= Mod;
		}

		printf("%d %lld\n", ans, cnt);
	}
	return 0;
}