#pragma GCC optimize("O3,unroll-loops")
#include<bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
typedef long long LL;
typedef unsigned long long ULL;
#define fi first
#define se second
#define all(x) ((x).begin()), ((x).end())
#define pb push_back
typedef double D;
const int N = 200011;
int tp[N], d[N], nxt[N];
pair<string, int> find(string s) {
    int i, j, k, l, n = s.length();
    s += s;
    for(i = 0, j = 1; j < n; ) {
        for(k = 0; k < n && s[i + k] == s[j + k]; k++);
        if(k >= n) break;
        if(s[i + k] < s[j + k]) {
            j += k + 1;
        }
        else l = i + k, i = j, j = max(l, j) + 1;
    }
    return make_pair(s.substr(i, n), i);
    
}
struct Op {
    int coe, delta;
    void print() const {
        printf("c = %d, d = %d\n", coe, delta);
    }
};
int per;
Op operator + (const Op & a, const Op & b) {
    return Op{a.coe * b.coe, (a.delta * b.coe % per + b.delta + per) % per};
}
bool operator < (const Op & a, const Op & b) {
    return a.coe < b.coe || a.coe == b.coe && a.delta < b.delta;
}
int main() {
    string s;
    cin >> s;
    int n = s.size();
    int k;
    scanf("%d", &k);
    for(int i = 0; i < k; i++) {
        static char st[11];
        scanf("%s", st);
        tp[i] = st[0];
        if(tp[i] != 'I') {
            scanf("%d", &d[i]);
            if(tp[i] == 'R') {
                d[i] = d[i] % n;
            }else {
                d[i] = (-d[i] + n) % n;
            }
        }
    }
    int revdelta = -1;
    string t = s;
    reverse(all(t));
    pair<string, int> ms = find(s), mt = find(t);
    if(ms.first == mt.first) {
        revdelta = ms.se - mt.se;
    }
    s = '$' + s;
    nxt[0] = -1;
    for(int i = 1; i <= n; i++) {
        int p = nxt[i - 1];
        while(p != -1 && s[p + 1] != s[i]) {
            //printf("p = %d\n", p);
            p = nxt[p];
        }
        nxt[i] = p + 1;
    }
    per = n - nxt[n];
    LL ans = 0;
    //printf("per = %d, revdelta = %d\n", per, revdelta);
    if(revdelta == -1) {
        Op cur{1, 0};
        map<Op, int> cnt;
        cnt[cur] = 1;
        for(int i = 0; i < k; i++) {
            Op opi;
            if(tp[i] == 'I') {
                opi = Op{-1, 0};
            } else {
                opi = Op{1, d[i] % per};
            }
            cur = cur + opi;
            //cur.print();
            ans += cnt[cur];
            cnt[cur]++;
        }
        /*for(int i = 0, j; i < k; i = j) {
            if(tp[i] == 'I') {
                j = i + 1;
                continue;
            }
            for(j = i; tp[j] != 'I'; j++);
            map<int, int> cnt;
            cnt[0] = 1;
            int cur = 0;
            for(int k = i; k < j; k++) {
                cur = (cur + d[k]) % per;
                ans += cnt[cur];
                cnt[cur]++;
            }
        }*/
    } else {
        /*if(per == 1) {
            ans = (k * (k + 1ll) / 2);
        } else */
        /*{
            map<int, int> cnt;
            cnt[0] = 1;
            int cur = 0;
            for(int i = 0; i < k; i++) {
                if(tp[i] == 'I') {
                    cur = (per - cur + revdelta) % per;
                } else {
                    cur = (cur + d[i]) % per;
                }
                ans += cnt[cur];
                cnt[cur]++;
            }
        }*/
        Op cur{1, 0};
        map<Op, int> cnt;
        cnt[cur] = 1;
        for(int i = 0; i < k; i++) {
            Op opi;
            if(tp[i] == 'I') {
                opi = Op{-1, revdelta};
            } else {
                opi = Op{1, d[i] % per};
            }
            cur = cur + opi;
            //cur.print();
            ans += cnt[cur];
            ans += cnt[Op{-cur.coe, (per - cur.delta) % per}];
            cnt[cur]++;
        }
    }
    printf("%lld\n", ans);
}