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 bool boolean;

#define ll long long

const ll Lim = 1e18;
const int Mod = 1e9 + 7;

#define pll pair<ll, ll>

int Q;
ll x, y;
vector<pll> fib;
vector<vector<pll>> F;

boolean check(pll x) {
	return x.first <= Lim && x.second <= Lim;
}

void prepare() {
	F.push_back({pll(2, 1)}); // x > y
	F.push_back({pll(3, 2), pll(4, 3)});
	while (!F.back().empty()) {
		vector<pll> g;
		boolean first = true;
		vector<pll>& f = F.back();
		for (auto x : f) {
			if (first) {
				pll y = pll(x.second + x.first, x.first);
				if (check(y))
					g.push_back(y);
				y = pll(x.second * 2 + x.first, x.first + x.second);
				if (check(y))
					g.push_back(y);
			} else {
				pll y = pll(x.second + x.first, x.first);
				if (check(y))
					g.push_back(y);
			}
			first = false;
		}
		F.push_back(g);
	}
	fib.emplace_back(2, 1);
	ll a = 2, b = 1;
	while (true) {
		b += a;
		swap(a, b);
		if (a <= Lim && b <= Lim) {
			fib.emplace_back(a, b);
		} else {
			break;
		}
	}
}

int M;
int work(ll n, ll m) {
	ll ans = 0;
	for (auto x : F[M]) {
		if (x.first <= n && x.second <= m) {
			ans = (ans + (n - x.first) / x.second + 1) % Mod;
		}
	}
	return ans;
}

int main() {
	prepare();
	scanf("%d", &Q);
	while (Q--) {
		scanf("%lld%lld", &x, &y);
		if (x < y)
			swap(x, y);
		if (x == 1 && y == 1) {
			M = 0;
		} else {
			int l = 0, r = (signed) fib.size() - 1, mid;
			while (l <= r) {
				mid = (l + r) >> 1;
				if (x >= fib[mid].first && y >= fib[mid].second) {
					l = mid + 1;
				} else {
					r = mid - 1;
				}
			}
			M = l - 1;
		}
		int ans = work(x, y);
		ans += work(y, x);
		if (M == 0)
			ans += min(x, y);
		printf("%d %d\n", M + 1, ans % Mod);
	}
	return 0;
}