Problem: Given a string s, process q queries, each having one of the following forms:

  * 1 i c — Change the i-th character in the string to c. 
  * 2 l r y — Consider the substring of s starting at position l and ending at position r. Output the number of times y occurs as a substring in it. 

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 105) of lowercase English letters.

The second line contains an integer q (1 ≤ q ≤ 105) — the number of queries to process.

The next q lines describe the queries and may have one of the following forms:

  * 1 i c (1 ≤ i ≤ |s|) 
  * 2 l r y (1 ≤ l ≤ r ≤ |s|) 



c is a lowercase English letter and y is a non-empty string consisting of only lowercase English letters.

The sum of |y| over all queries of second type is at most 105.

It is guaranteed that there is at least one query of second type.

All strings are 1-indexed.

|s| is the length of the string s.

Output

For each query of type 2, output the required answer in a separate line.

Examples

Input

ababababa
3
2 1 7 aba
1 5 c
2 1 7 aba


Output

3
1


Input

abcdcbc
5
2 1 7 bc
1 4 b
2 4 7 bc
1 2 a
2 1 4 aa


Output

2
2
1

Note

Consider the first sample case. Initially, the string aba occurs 3 times in the range [1, 7]. Note that two occurrences may overlap. 

After the update, the string becomes ababcbaba and now aba occurs only once in the range [1, 7].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int dr[]{-1, -1, 0, 1, 1, 1, 0, -1};
const int dc[]{0, 1, 1, 1, 0, -1, -1, -1};
void run() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int sqrtQ = 100;
struct suffix_automaton {
  struct state {
    int len, link = 0, cnt = 0;
    bool terminal = false, is_clone = false;
    vector<int> nxt;
    state(int len = 0) : len(len), nxt(26, -1) {}
    bool have_next(char ch) { return nxt[ch - 'a'] != -1; }
    void clone(const state &other, int nlen) {
      len = nlen;
      nxt = other.nxt;
      link = other.link;
      is_clone = true;
    }
  };
  vector<state> st;
  int last = 0;
  suffix_automaton() {
    st.push_back(state());
    st[0].link = -1;
  }
  suffix_automaton(const string &s) : suffix_automaton() {
    for (char ch : s) extend(ch);
    calc_number_of_occurrences();
  }
  void extend(char c) {
    int cur = st.size();
    st.push_back(state(st[last].len + 1));
    st[cur].cnt = 1;
    int p = last;
    last = cur;
    while (p != -1 && !st[p].have_next(c)) {
      st[p].nxt[c - 'a'] = cur;
      p = st[p].link;
    }
    if (p == -1) return;
    int q = st[p].nxt[c - 'a'];
    if (st[p].len + 1 == st[q].len) {
      st[cur].link = q;
      return;
    }
    int clone = st.size();
    st.push_back(state());
    st[clone].clone(st[q], st[p].len + 1);
    while (p != -1 && st[p].nxt[c - 'a'] == q) {
      st[p].nxt[c - 'a'] = clone;
      p = st[p].link;
    }
    st[q].link = st[cur].link = clone;
  }
  void calc_number_of_occurrences() {
    vector<vector<int>> lvl(st[last].len + 1);
    for (int i = 1; i < st.size(); i++) lvl[st[i].len].push_back(i);
    for (int i = st[last].len; i >= 0; i--)
      for (auto cur : lvl[i]) st[st[cur].link].cnt += st[cur].cnt;
  }
  int count(const string &y) {
    int cur = 0;
    for (auto &it : y) {
      if (!st[cur].have_next(it)) return 0;
      cur = st[cur].nxt[it - 'a'];
    }
    return st[cur].cnt;
  }
} SA[(int)1e4];
string y;
vector<int> longestPrefix;
vector<int> Fail[26];
int fail(int k, char nxt) {
  int &rt = Fail[nxt - 'a'][k];
  if (~rt) return rt;
  while (k > 0 && y[k] != nxt) k = longestPrefix[k - 1];
  if (nxt == y[k]) k++;
  return rt = k;
}
void failure_function() {
  int n = y.size();
  longestPrefix = vector<int>(n);
  for (int i = 0; i < 26; i++) Fail[i] = vector<int>(n + 1, -1);
  for (int i = 1, k = 0; i < n; i++) longestPrefix[i] = k = fail(k, y[i]);
}
int main() {
  run();
  string s;
  int q;
  cin >> s >> q;
  for (int i = 0; i < (int)(s.size()); i += sqrtQ)
    SA[i / sqrtQ] = suffix_automaton(s.substr(i, sqrtQ));
  auto solve = [&](int l, int r) {
    int match = 0, cnt = 0;
    for (int i = l; i <= r; i++) {
      match = fail(match, s[i]);
      if (match == (int)(y.size())) {
        cnt++;
        match = longestPrefix[match - 1];
      }
    }
    return cnt;
  };
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int idx;
      char ch;
      cin >> idx >> ch;
      idx--;
      s[idx] = ch;
      SA[idx / sqrtQ] = suffix_automaton(s.substr(idx / sqrtQ * sqrtQ, sqrtQ));
    } else {
      int l, r;
      cin >> l >> r >> y;
      l--, r--;
      failure_function();
      if ((int)(y.size()) >= sqrtQ || r - l <= 2 * sqrtQ) {
        cout << solve(l, r) << "\n";
        continue;
      }
      int ans = 0, len = (int)(y.size());
      int st = l / sqrtQ, ed = r / sqrtQ;
      ans += solve(l, (st + 1) * sqrtQ + len - 2);
      ans += solve(ed * sqrtQ - len + 1, r);
      for (int i = st + 1; i < ed; i++) ans += SA[i].count(y);
      for (int i = st + 2; i < ed; i++)
        ans += solve(i * sqrtQ - len + 1, i * sqrtQ + len - 2);
      cout << ans << "\n";
    }
  }
}