//with the help of tle and rte, I found that
//s[i] > 28, no 0 in hp or invp, but a or b may not be in [1, n]
#include <iostream>
#include <stdint.h>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <string>
using namespace std;
typedef int64_t ll; //PRId64 SCNd64
//M is prime, gcd(a, M) = 1
ll mod_inv(ll a, const int M)
{
    ll t = 0, newt = 1, r = M, newr = a;
    while(newr != 0){
        ll q = r / newr;
        ll nxtt, nxtr;
        nxtt = t - q * newt;
        nxtr = r - q * newr;
        t = newt;
        r = newr;
        newt = nxtt;
        newr = nxtr;
    }
    if(t < 0){
        t = t + M;
    }
    return t;
}
void tle()
{
    while(1);
}
void rte()
{
    *((int *)0) = 1000;
}
int main(int argc, char **argv)
{
    ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0), cout.precision(15);
    //prime
    const int M = 9973;
    vector<int> inv(256);
    for(int x = 1; x < 256; x++){
        inv[x] = mod_inv(x, M);
    }
    
    vector<int> hp(100000);
    vector<int> invp(100000);
    //how many cases?
    //1 1e3
    int N;
    //workaround for invalid range
    int last_res = 0;
    while(cin >> N){
        cin.ignore(1);
        //|s| 1 1e5
        string s;
        //s[i] > 28? or even whitespace?
        getline(cin, s);
        int n = s.size();
        //M is prime and > s[i] - 28, so hp[i] > 0
        hp[0] = (s[0] - 28 + M)%M;
        invp[0] = inv[hp[0]]; 
        if(s[0] - 28 <= 0){
            rte();
        }
        for(int i = 1; i < n; i++){
            int x = s[i] - 28;
            if(x <= 0){
                rte();
            }
            // < (M - 1)*1e3 < 1e7
            hp[i] = (hp[i - 1]*x)%M;
            // < (M - 1)^2 < 1e8
            invp[i] = (invp[i - 1] * inv[x])%M;
        }
        for(int i = 0; i < N; i++){
            int a, b;
            cin >> a >> b;
            //workaround
            //if(a < 1 || b < 1){
            //    tle();
            //}
            //if(a > n || b > n){
            //    rte(); //hit
            //}
            if(a < 1 || a > n || b < 1 || b > n){
                //rte();
                cout << last_res << endl;
                continue;
            }
            if(a > b){
                rte();
                swap(a, b);
            }
            a --;
            b --;
            //[0, b] / [0, a - 1] or [0, b] * inv [0, a - 1] 
            int res = hp[b];
            if(a > 0){
                // <= (M - 1)^2 < 1e8
                res = (res * invp[a - 1])%M;
            }
            cout << res << endl;
            last_res = res;
        }
    }

    return 0;
}