/*

此题要求 (G^1+G^2+...+G^t2)-(G^1+G^2+...+G^(t1-1))

求和的方法是再次二分,k=6时

G + G^2 + G^3 + G^4 + G^5 + G^6 = G + G^2 + G^3 + G^3 * (G + G^2 + G^3) = (1 + G^3) * (G + G^2 + G^3)

这样计算可以使k的规模减少一半，快速幂求出G^3后可递归计算G+G^2+G^3,即得答案

*/

#include <iostream>

#include <cstdio>

#include <cstring>

#include <map>

using namespace std;

#define MOD 2008

#define Mat 35 //矩阵大小

struct mat{//矩阵结构体，a表示内容，r行c列 矩阵从1开始

    int a[Mat][Mat];

    int r, c;

    mat() {

        r = c = ;

        memset(a, , sizeof(a));

    }

};  

void print(mat m) {

    //printf("%d\n", m.size);

    for(int i = ; i < m.r; i++) {

        for(int j = ; j < m.c; j++) printf("%d ", m.a[i][j]);

        putchar('\n');

    }

}  

mat mul(mat m1, mat m2, int mod) {

    mat ans  = mat();

    ans.r = m1.r, ans.c = m2.c;

    for(int i = ; i <= m1.r; i++)

        for(int j = ; j <= m2.r; j++)

            if(m1.a[i][j])

                for(int k = ; k <= m2.c; k++)

                    ans.a[i][k] = (ans.a[i][k] + m1.a[i][j] * m2.a[j][k]) % mod;

    return ans;

}

mat add(mat m1, mat m2, int mod) {

    mat ans  = mat();

    ans.r = ans.c = m1.r;

    for(int i = ; i <= m1.r; i++)

        for(int j = ; j <= m1.r; j++)

            ans.a[i][j] = (m1.a[i][j] + m2.a[i][j]) % mod;

    return ans;

}    

mat quickmul(mat m, int n, int mod) {

    mat ans = mat();

    for(int i = ; i <= m.r; i++) ans.a[i][i] = ;

    ans.r = m.r, ans.c = m.c;

    while(n) {

        if(n & ) ans = mul(m, ans, mod);

        m = mul(m, m, mod);

        n >>= ;

    }

    return ans;

}  

mat A;

mat sum(mat m, int n, int mod) {   //m^1+m^2+...+m^n

    if(n == ) return m;

    if(n & ) return add(quickmul(m, n, mod), sum(m, n-, mod), mod);             //是否加m^n

    else return mul(add(quickmul(m, n>>, mod), A, mod), sum(m, n>>, mod), mod); // (1 + m^(n/2)) * (m + m^2 +...+ m^(n/2))

}

/*

初始化ans矩阵

mat ans = mat();

ans.r = R, ans.c = C;

ans = quickmul(ans, n, mod);

*/

int main() {

    A = mat();

    for(int i = ; i <= ; i++) A.a[i][i] = ;

    int n;

    while(~scanf("%d", &n)) {

        mat G = mat();

        G.r = G.c = ;

        map <int, int> mp;

        int rank = ;

        while(n--) {

            int a, b;

            scanf("%d%d", &a, &b);

            if(!mp[a]) mp[a] = rank++;

            if(!mp[b]) mp[b] = rank++;

            G.a[mp[a]][mp[b]]++;

        }

        int k;

        scanf("%d", &k);

        for(int i = ; i < k; i++) {

            int v1, v2, t1, t2;

            scanf("%d%d%d%d", &v1, &v2, &t1, &t2);

            v1 = mp[v1];

            v2 = mp[v2];

            if(t1 > t2) swap(t1, t2);

            if(!t1){

                if(!t2) puts("");

                else {

                    int ans = sum(G, t2, MOD).a[v1][v2];

                    printf("%d\n", ans);

                }

            }

            else if(t1 == ) {

                int ans = sum(G, t2, MOD).a[v1][v2];

                printf("%d\n", ans);

            }

            else {

                int ans1 = sum(G, t1-, MOD).a[v1][v2];

                int ans2 = sum(G, t2, MOD).a[v1][v2];

                printf("%d\n", (ans2 - ans1 + MOD) % MOD);

            }

        }

    }

    return ;

}