#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#define QWQ cout << "QwQ" << endl;
#define ll long long
#include <vector>
#include <queue>
#include <stack>
#include <map>
#define ls now << 1
#define rs now << 1 | 1
using namespace std;
const int N = 401010;
const int qwq = 4003030;
const int inf = 0x3f3f3f3f;
int n, m;
int cnt;
char s[N];
struct T {
    int to[28], faith, from;
} f[N];             // trie树
vector<int> d[N];   // fail指针树
vector<int> at[N];  // tire树上点是哪些串的结束。
vector<int> wen[N]; //这个点（即y串）有哪些问题。。.
vector<int> e[N];   //原来的树，因为我BFS时将原来的trie树改成了trie图，所以我需要记录最开始时的树。
queue<int> q;
int wei[N];            //此字符串结束位置。
int askman[N], ans[N]; //此问题：谁发起问题（即x串），问题答案；
int tree[qwq], id[N], siz[N];

void BFS() {
    for (int i = 1; i <= 26; i++)
        if (f[0].to[i]) q.push(f[0].to[i]);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        for (int i = 1; i <= 26; i++) {
            int v = f[u].to[i];
            if (v) {
                f[v].faith = f[f[u].faith].to[i];
                q.push(v);
            } else
                f[u].to[i] = f[f[u].faith].to[i];
        }
    }
    for (int i = 1; i <= cnt; i++) d[f[i].faith].push_back(i); //建fail树。
}

void DFS1(int u) { //遍历fail树，算出每个点子树大小，并给每个点一个编号。
    siz[u] = 1;
    id[u] = ++cnt;
    for (int i = 0; i < d[u].size(); i++) {
        int v = d[u][i];
        DFS1(v);
        siz[u] += siz[v];
    }
}

void insert(int now, int l, int r, int we, int g) { //单点修改
    if (l == r) {
        tree[now] += g;
        return;
    }
    int mid = (l + r) >> 1;
    if (we <= mid)
        insert(ls, l, mid, we, g);
    else
        insert(rs, mid + 1, r, we, g);
    tree[now] = tree[ls] + tree[rs];
}

int query(int now, int l, int r, int x, int y) { //区间查询
    if (x <= l && r <= y) { return tree[now]; }
    int mid = l + r >> 1, res = 0;
    if (x <= mid) res += query(ls, l, mid, x, y);
    if (y > mid) res += query(rs, mid + 1, r, x, y);
    return res;
}

void TREE(int now) {               //遍历原树
    insert(1, 1, cnt, id[now], 1); //每到一个点就加上这个点的贡献
    for (int i = 0; i < at[now].size(); i++) {
        int u = at[now][i];
        for (int j = 0; j < wen[u].size(); j++) {
            int qu = wen[u][j], v = wei[askman[qu]]; // qu为问题编号，v为串x的结尾点
            ans[qu] = query(1, 1, cnt, id[v], id[v] + siz[v] - 1);
        }
    }
    for (int i = 0; i < e[now].size(); i++) TREE(e[now][i]);
    insert(1, 1, cnt, id[now], -1);
}

int main() {
    int x, y;
    scanf("%s", s);
    int len = strlen(s), now = 0;
    for (int i = 0; i < len; i++) {
        int v = s[i] - 'a' + 1;
        if (s[i] == 'P') {
            at[now].push_back(++n);
            wei[n] = now;
            continue;
        }
        if (s[i] == 'B') {
            now = f[now].from;
            continue;
        }
        if (!f[now].to[v]) f[now].to[v] = ++cnt, e[now].push_back(f[now].to[v]);
        f[f[now].to[v]].from = now;
        now = f[now].to[v]; //涉及到退格，需要记录from。
    }
    BFS();
    cnt = 0;
    DFS1(0);
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &askman[i], &x);
        wen[x].push_back(i);
    }
    TREE(0);
    for (int i = 1; i <= m; i++)
        printf("%d\n", ans[i]);
    return 0;
}