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>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cout << #x << " = " << (x) << endl

const ll MOD = 1000000007;
const ll MX = 1000000000000000000;

vector<ll> fib; //87
vector<pair<ll,ll>> cand[100];

void gen() {
	for (int i = 1; i < (int)fib.size() - 1; ++i) {
		cand[i].eb(fib[i], fib[i+1]);

		for (int p = 1; p < i; ++p) {
			ll x = 1, y = 1;
			for (ll j = 1; j <= i; ++j) {
				ll nx, ny;
				if (j != p) {
					nx = y;
					ny = x + y;
				} else {
					nx = y;
					ny = x + y * 2;
				}
				x = nx; y = ny;
			}
			cand[i].eb(x, y);
		}
	}
}

int main() {
	fib.pb(1); fib.pb(1);
	for (int i = 2; ; ++i) {
		ll x = fib[i-2] + fib[i-1];
		fib.pb(x);
		if (x > MX) {
			break;
		}
	}

	gen();

	int Q; cin >> Q;
	while (Q--) {
		ll X, Y;
		cin >> X >> Y;
		if (X > Y) swap(X, Y);

		int maxi = 1;
		
		if (X != 1) {
			for (int i = 0; i < (int)fib.size()-1; ++i) {
				if (fib[i] <= X && fib[i + 1] <= Y) {
					maxi = i;
				}
			}
		}

		ll ans = 0;
		if (maxi == 1) {
			ans = X % MOD * Y % MOD;
		} else {
			for (auto p : cand[maxi]) {
				ll a = p.fi, b = p.se;
				ll cnt = 0;
				if (a <= X && b <= Y) {
					cnt += (Y - b) / a + 1;
				}
				if (b <= X && a <= Y) {
					cnt += (Y - a) / b + 1;
				}

				ans += cnt % MOD;
				if (ans >= MOD) ans -= MOD;
			}
		}

		cout << maxi << " " << ans << "\n";
	}

	return 0;
}