F(x)，相减（数位DP，减法的艺术，状态设计）

//http://acm.hdu.edu.cn/showproblem.php?pid=4734
#include <bits/stdc++.h>
#define ll long long
#define all(a) (a).begin(), (a).end()
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
using namespace std;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1e3 + 10;

int num[20], k;
ll f[20][4600];//最多还能加多少权重

ll dfs(int pos, int mx, bool lead, bool limit)
{
    if(pos == 0) return 1; 
    if(!lead && !limit && f[pos][mx] != -1) return f[pos][mx];
    ll ans = 0;
    int up = limit ? num[pos] : 9;
    for(int i = 0; i <= up; i++)
    {
        if(mx >= (1 << (pos - 1)) * i)
            ans += dfs(pos - 1, mx - (1 << (pos - 1)) * i, lead && i == 0, limit && i == up);
        else 
            break;
    }
    if(!lead && !limit) f[pos][mx] = ans;
    return ans;
}

ll run(ll x)
{
    int len = 0;
    while(x)
    {
        num[++len] = x % 10;
        x /= 10;
    }
    return dfs(len, k, true, true);
}

int tot;
void solve()
{
    ll A, B;
    cin >> A >> B;
    
    k = 0;
    for(int i = 0; A; i++, A /= 10) k += (1 << i) * (A % 10);

    cout << "Case #" << ++tot << ": " << run(B) << '\n'; 
}

signed main()
{
    IOS;
    int t = 1;
    memset(f, -1, sizeof f);
    cin >> t;
    while (t--)
        solve();
    return 0;
}