数位
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long i64;
int len;
int nums[20];
i64 f[20][10];
i64 dfs(int pos, int mod, int limit) {
if (!pos) return !!mod;
if (!limit && ~f[pos][mod]) return f[pos][mod];
int up = limit ? nums[pos] : 9; i64 res = 0;
for (int i = 0; i <= up; i ++ )
if (i != 9) res += dfs(pos - 1, (mod + i) % 9, limit && i == up);
return limit ? res : f[pos][mod] = res;
}
i64 calc(i64 x) {
len = 0;
memset(f, -1, sizeof f);
while (x) nums[ ++ len] = x % 10, x /= 10;
return dfs(len, 0, 1);
}
void solve(int C) {
i64 l, r;
scanf("%lld%lld", &l, &r);
printf("Case #%d: %lld\n", C, calc(r) - calc(l - 1));
}
int main() {
int _, C = 0; scanf("%d", &_);
while (_ -- ) solve( ++ C);
return 0;
}