#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}
 
void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf double
#define ll long long 

const int maxn = 110;
const int inf = 1e9;
const lf eps = 1e-8;
const int mod = 1e9+7;

#define pii pair<int,int >
#define fr first
#define sc second

#define vec vector<pii >
#define mp make_pair
#define pb push_back

#define iter vector <pii > :: iterator

vec t[maxn];
int n,m,q,f[maxn];

signed main() {
    t[1].pb(mp(1,2)),t[1].pb(mp(1,3)),t[1].pb(mp(1,4));
    f[0]=f[1]=1;for(int i=2;i<=100;i++) f[i]=f[i-1]+f[i-2];
    for(int i=1;i<=100;i++) 
        for(int j=0,l=t[i].size()-1;j<=l;j++) {
            int x=t[i][j].sc,y=t[i][j].fr+x;
            while(y<=f[i+3]+f[i]) t[i+1].pb(mp(x,y)),y+=x;
        }read(q);
    for(int i=1;i<=q;i++) {
        read(n),read(m);if(n>m) swap(n,m);
        int p=1,ans=0;
        while(f[p+1]<=n&&f[p+2]<=m) p++;
        printf("%lld ",p);
        if(p==1) {write(n%mod*m%mod);continue;}
        for(int j=0,l=t[p-1].size()-1;j<=l;j++) {
            int x=t[p-1][j].fr,y=t[p-1][j].sc;//printf("debug :: %d %d\n",x,y);
            if(y<=n) ans=(ans+(m-x)/y)%mod;
            if(y<=m) ans=(ans+(n-x)/y)%mod;
        }write(ans);
    }
    return 0;
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}
 
void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf double
#define ll long long 

const int maxn = 110;
const int inf = 1e9;
const lf eps = 1e-8;
const int mod = 1e9+7;

#define pii pair<int,int >
#define fr first
#define sc second

#define vec vector<pii >
#define mp make_pair
#define pb push_back

#define iter vector <pii > :: iterator

vec t[maxn];
int n,m,q,f[maxn];

signed main() {
    t[1].pb(mp(1,2)),t[1].pb(mp(1,3)),t[1].pb(mp(1,4));
    f[0]=f[1]=1;for(int i=2;i<=100;i++) f[i]=f[i-1]+f[i-2];
    for(int i=1;i<=100;i++) 
        for(int j=0,l=t[i].size()-1;j<=l;j++) {
            int x=t[i][j].sc,y=t[i][j].fr+x;
            while(y<=f[i+3]+f[i]) t[i+1].pb(mp(x,y)),y+=x;
        }read(q);
    for(int i=1;i<=q;i++) {
        read(n),read(m);if(n>m) swap(n,m);
        int p=1,ans=0;
        while(f[p+1]<=n&&f[p+2]<=m) p++;
        printf("%lld ",p);
        if(p==1) {write(n%mod*m%mod);continue;}
        for(int j=0,l=t[p-1].size()-1;j<=l;j++) {
            int x=t[p-1][j].fr,y=t[p-1][j].sc;//printf("debug :: %d %d\n",x,y);
            if(y<=n) ans=(ans+(m-x)/y)%mod;
            if(y<=m) ans=(ans+(n-x)/y)%mod;
        }write(ans);
    }
    return 0;
}

```