#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define lowbit(x) (x & (-x))const double eps = 1e-9;inline int dcmp(double a, double b){    if(fabs(a - b) < eps) return 0;    return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 20 + 10;const int MAXT = 10000 + 10;using namespace std;int digit[MAXN];LL dp[MAXN][2][MAXN];LL dfs(int pos, int pre, int len, bool leadingzero, bool limit){//pos--当前位，pre--更高一位的数是奇还是偶，len--连续长度，leadingzero--是否有前导零，limit--当前位的数字是否有限制    if(!pos) return (pre & 1) != (len & 1);    if(!limit && dp[pos][pre][len] != -1) return dp[pos][pre][len];    LL ans = 0;    int up = limit ? digit[pos] : 9;    for(int i = 0; i <= up; ++i){        if(leadingzero){            if(i == 0){                ans += dfs(pos - 1, 0, 0, true, limit && i == up);            }            else{                ans += dfs(pos - 1, i & 1, 1, false, limit && i == up);            }        }        else{            if(i & 1){                if(pre & 1){                    ans += dfs(pos - 1, i & 1, len + 1, false, limit && i == up);                }                else{                    if(len & 1){//若当前位是奇数，前一位是偶数，且已经有奇数长度的偶数，则可以继续延伸                        ans += dfs(pos - 1, i & 1, 1, false, limit && i == up);                    }                }            }            else{                if(pre & 1){                    if(!(len & 1)){                        ans += dfs(pos - 1, i & 1, 1, false, limit && i == up);                    }                }                else{                    ans += dfs(pos - 1, i & 1, len + 1, false, limit && i == up);                }            }        }    }    if(!limit) dp[pos][pre][len] = ans;    return ans;}LL solve(LL x){    int cnt = 0;    while(x){        digit[++cnt] = x % 10;        x /= 10;    }    return dfs(cnt, 0, 0, true, true);}int main(){    int T;    scanf("%d", &T);    int kase = 0;    memset(dp, -1, sizeof dp);    while(T--){        LL L, R;        scanf("%lld%lld", &L, &R);        printf("Case #%d: %lld\n", ++kase, solve(R) - solve(L - 1));    }    return 0;}