#define _CRT_SECURE_NO_WARNINGS 1
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define double long double
//#define int long long
#define lc u<<1
#define rc u<<1|1
#define endl "\n"
#define X first
#define Y second
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> PII;
const ll N = 1e6 + 50;
const ll maxm = 2e18;
const ll mod = 998244353;

ll exgcd(ll a, ll b, ll& x, ll& y);

ll dfs(ll x) {
    if (x == 0)return 1;
    if (x > 0 && x % 3 == 0)return dfs(x / 3) + 1;
    if (x > 0 && x % 3 != 0)return dfs(x - 1) + 1;
}

ll base[50];

ll ksm(ll a, ll b) {
    ll base = a;
    ll ans = 1;
    while (b) {
        if (b & 1)ans *= base;
        base *= base;
        b >>= 1;
    }
    return ans;
}

void solve()
{
    //复盘一下，化为三进制后，我们观察所有第一次出现的元素的三进制发现，
    //他比前置元素都大，而他的三进制最后的几位全部是2，猜测，越放2越优
    //观察题目中给的式子，在三进制尺度下，我们发现，三进制每个位置上的数字越大
    //获取操作三的次数越多，加1越多，第二个操作本质是右移操作
    //除三操作，这意味着我们一直向前找比r小的元素，然后再某一位减1
    //对后面全赋二，这样就找到更优的一种情况，

    //cout << ksm(3, 50) << endl;
    ll l, r; cin >> l >> r;
    for (int i = 0; i <= 49; i++)base[i] = 0;
    ll idx = -1, mid = r;
    while (mid) {//完，先得到的是最高位
        base[++idx] = mid % 3;
        mid /= 3;
    }
    ll ans = dfs(r); ll cnt = 0;
    for (ll i = idx; i >= 0; i--) {
        ll mid = r;
        if (base[i] == 0) {
            cnt++;
            continue;
        }
        mid -= ksm(3, cnt);
        for (ll j = 0; j < cnt; j++) {
            mid += 2 * ksm(3, j);
        }
        ans = max(ans, dfs(mid));
        cnt++;
    }
    cout << ans << endl;
    return;
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    ll t = 1; cin >> t;
    while (t--)solve();
    return 0;
}
/*PonyHex*/


ll exgcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
        x = 1; y = 0;
        return a;
    }
    ll g = exgcd(b, a % b, x, y);
    ll temp = x;
    x = y;
    y = temp - (a / b) * y;
    return g;
}
/*
1
11
1 2 3 2 5 6 7 6 9 1
*/
/*
ll ksm(ll a, ll b) {
    ll base = a;
    ll ans = 1;
    while (b) {
        if (b & 1)ans *= base;
        base *= base;
        b >>= 1;
    }
    return ans;
}*/
/*
ll gcd(ll a, ll b) {
    return b ? gcd(b, a % b) : a;
}*/