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 int mod = 1e9 + 7;
void prime(long long int a, vector<long long int> &v) {
  for (long long int i = 2ll; i * i <= a; i++) {
    if (a % i == 0ll) v.push_back(i);
    while (a % i == 0ll) {
      a = a / i;
    }
  }
  if (a > 1ll) v.push_back(a);
}
bool isprime(long long int a) {
  for (long long int i = 2; i * i <= a; i++)
    if (a % i == 0) return false;
  return true;
}
long long int expmod(long long int a, long long int b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long int m1 = expmod(a, b / 2) % mod;
  if (b % 2) return ((m1 * m1) % mod * a % mod) % mod;
  return (m1 * m1) % mod;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
struct myComp {
  constexpr bool operator()(pair<int, pair<int, int>> const &a,
                            pair<int, pair<int, int>> const &b) const noexcept {
    if (a.first < b.first) return true;
    if (a.first > b.first) return false;
    return a.second.first > b.second.first;
  }
};
class compare {
 public:
  bool operator()(pair<int, long long int> &p, pair<int, long long int> &q) {
    return p.second >= q.second;
  }
};
long long int bs1(long long int a[], long long int n, long long int x) {
  long long int anss = -1;
  long long int l = 0;
  long long int h = n - 1;
  while (l <= h) {
    long long int mid = l + (h - l) / 2;
    if (a[mid] <= x) {
      anss = mid;
      l = mid + 1;
    } else
      h = mid - 1;
  }
  return anss;
}
long long int bs2(long long int a[], long long int n, long long int x) {
  long long int anss = -1;
  long long int l = 0;
  long long int h = n - 1;
  while (l <= h) {
    long long int mid = l + (h - l) / 2;
    if (a[mid] > x) {
      anss = mid;
      h = mid - 1;
    } else
      l = mid + 1;
  }
  return anss;
}
int bs3(vector<int> a, int x) {
  int anss = -1;
  int l = 0;
  int h = a.size() - 1;
  while (l <= h) {
    int mid = l + (h - l) / 2;
    if (a[mid] <= x) {
      anss = mid;
      l = mid + 1;
    } else
      h = mid - 1;
  }
  return anss;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int test;
  test = 1;
  while (test--) {
    string s;
    cin >> s;
    int n = s.size();
    vector<set<int>> v(27);
    for (int i = 0; i < n; i++) {
      v[s[i] - 'a'].insert(i);
    }
    int q;
    cin >> q;
    for (int i = 0; i < q; i++) {
      int x;
      cin >> x;
      if (x == 1) {
        int j;
        char f;
        cin >> j >> f;
        j--;
        v[s[j] - 'a'].erase(j);
        s[j] = f;
        v[f - 'a'].insert(j);
      } else {
        int c = 0;
        int l, r;
        cin >> l >> r;
        l--;
        r--;
        for (int i = 0; i < 26; i++) {
          auto it = v[i].lower_bound(l);
          if (it != v[i].end() && *it <= r) {
            c++;
          }
        }
        cout << c << "\n";
      }
    }
    cout << "\n";
  }
}
