You are given a string s consisting of lowercase Latin letters and q queries for this string.

Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".

There are two types of queries: 

  * 1~ pos~ c (1 ≤ pos ≤ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c); 
  * 2~ l~ r (1 ≤ l ≤ r ≤ |s|): calculate the number of distinct characters in the substring s[l; r]. 

Input

The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.

The second line of the input contains one integer q (1 ≤ q ≤ 10^5) — the number of queries.

The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.

Output

For each query of the second type print the answer for it — the number of distinct characters in the required substring in this query.

Examples

Input


abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7


Output


3
1
2


Input


dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11


Output


5
2
5
2
6
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long fact(long long n) {
  long long res = 1;
  for (long long z = 2; z <= n; z++) res = res * z;
  return res;
}
long long Pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
long long amax(long long a[], long long n) {
  long long max = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] > max) max = a[i];
  }
  return max;
}
long long amin(long long a[], long long n) {
  long long min = 99999999999999999;
  for (long long i = 0; i < n; i++) {
    if (a[i] < min) min = a[i];
  }
  return min;
}
void ip(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
void op(long long a[], long long n) {
  for (long long i = 0; i < n; i++) {
    if (i == n - 1)
      cout << a[i] << endl;
    else
      cout << a[i] << " ";
  }
}
bool palindrome(long long le, long long ri, string s) {
  if (ri - le <= 0) return true;
  if (s[le] != s[ri]) return false;
  return palindrome(le + 1, ri - 1, s);
}
long long isprm(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
bool psqr(long double x) {
  long double sr = sqrt(x);
  return ((sr - floor(sr)) == 0);
}
int n;
const int man = 1 + 1e5;
char s[man];
int a[man];
int q;
set<int> S[man];
int main() {
  cin >> s;
  n = strlen(s);
  for (int i = 0; i < n; ++i) a[i] = s[i] - 'a';
  for (int i = 0; i < n; ++i) S[a[i]].insert(i);
  cin >> q;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int u;
      char c;
      cin >> u >> c;
      u--;
      int ret = c - 'a';
      if (a[u] == ret) continue;
      S[a[u]].erase(u);
      a[u] = ret;
      S[ret].insert(u);
    } else {
      int l, r;
      cin >> l >> r;
      int ans = 0;
      for (int i = 0; i < 26; ++i) {
        if (S[i].lower_bound(l - 1) != S[i].lower_bound(r)) ans++;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
