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

using namespace std;

typedef long long ll;
typedef pair<ll,ll> pll;

const ll inf=1000000000000000000ll;
const int mod=1000000007;

pll a[200][200];

int main()
{
    //freopen("input.txt","r",stdin);
    int l=1;a[1][1]=make_pair(1,2);a[1][2]=make_pair(1,3);
    while (a[l][1].second<=inf)
    {
        for (int i=1;i<=l+1;i++) a[l+1][i]=make_pair(a[l][i].second,a[l][i].first+a[l][i].second);
        ++l;a[l][l+1]=make_pair(a[l-1][1].second,a[l-1][1].second*2+a[l-1][1].first);
    }
    int q;scanf("%d",&q);
    while (q--)
    {
        ll x,y;scanf("%lld%lld",&x,&y);
        if (x>y) swap(x,y);
        if (x==y&&x==1) {puts("1 1");continue;}
        int ans1=0;
        for (int i=1;i<=l;i++)
            if (a[i][1].first>x||a[i][1].second>y) {ans1=i-1;break;}
        if (ans1==1) {printf("1 %d\n",x*y%mod);continue;}
        int ans2=0;
        for (int i=1;i<=ans1;i++)
        {
            ll u=a[ans1-1][i].first,v=a[ans1-1][i].second;
            if (v<=x) ans2=((y-u)/v+(x-u)/v+ans2)%mod;
        }
        printf("%d %d\n",ans1,ans2);
    }
    return 0;
}