#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
long long revrsno(long long n) {
  long long k = 0;
  vector<long long> z;
  while (n > 0) {
    z.push_back(n % 10);
    n /= 10;
  }
  for (long long i = 0; i < z.size(); i++) k = k * 10 + z[i];
  return k;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long powm(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 divide(long long a, long long b) {
  return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long Totfun(long long n) {
  long long z = n;
  if (n % 2 == 0) {
    while (n % 2 == 0) {
      n /= 2;
    }
    z /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (isPrime(i) && n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      z -= z / i;
    }
  }
  if (n > 1) {
    z -= z / n;
  }
  return z;
}
long long maxPrimeFactor(long long n) {
  long long maxPrime = -1;
  while (n % 2 == 0) {
    maxPrime = 2;
    n >>= 1;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      maxPrime = i;
      n = n / i;
    }
  }
  if (n > 2) maxPrime = n;
  return maxPrime;
}
long long fact[100001];
long long nCr(long long n, long long r) {
  if (r > n || r < 0) return 0;
  long long z = fact[n];
  z = mul(z, powm(fact[n - r], 1000000007 - 2));
  z = mul(z, powm(fact[r], 1000000007 - 2));
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    string s;
    cin >> s;
    long long q;
    cin >> q;
    vector<set<long long> > alpha(26);
    for (long long i = 0; i < s.size(); i++)
      alpha[int(s[i]) - 'a'].insert(i + 1);
    while (q--) {
      long long a;
      cin >> a;
      if (a == 1) {
        long long pos;
        char c;
        cin >> pos >> c;
        alpha[int(s[pos - 1]) - 'a'].erase(pos);
        s[pos - 1] = c;
        alpha[int(c) - 'a'].insert(pos);
      } else {
        long long l, r;
        cin >> l >> r;
        long long ans = 0;
        for (long long i = 0; i < 26; i++) {
          if (*alpha[i].lower_bound(l) >= l && *alpha[i].lower_bound(l) <= r &&
              *alpha[i].lower_bound(l) > 0)
            ans++;
        }
        cout << ans << endl;
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
long long revrsno(long long n) {
  long long k = 0;
  vector<long long> z;
  while (n > 0) {
    z.push_back(n % 10);
    n /= 10;
  }
  for (long long i = 0; i < z.size(); i++) k = k * 10 + z[i];
  return k;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long powm(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 divide(long long a, long long b) {
  return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long Totfun(long long n) {
  long long z = n;
  if (n % 2 == 0) {
    while (n % 2 == 0) {
      n /= 2;
    }
    z /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (isPrime(i) && n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      z -= z / i;
    }
  }
  if (n > 1) {
    z -= z / n;
  }
  return z;
}
long long maxPrimeFactor(long long n) {
  long long maxPrime = -1;
  while (n % 2 == 0) {
    maxPrime = 2;
    n >>= 1;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      maxPrime = i;
      n = n / i;
    }
  }
  if (n > 2) maxPrime = n;
  return maxPrime;
}
long long fact[100001];
long long nCr(long long n, long long r) {
  if (r > n || r < 0) return 0;
  long long z = fact[n];
  z = mul(z, powm(fact[n - r], 1000000007 - 2));
  z = mul(z, powm(fact[r], 1000000007 - 2));
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    string s;
    cin >> s;
    long long q;
    cin >> q;
    vector<set<long long> > alpha(26);
    for (long long i = 0; i < s.size(); i++)
      alpha[int(s[i]) - 'a'].insert(i + 1);
    while (q--) {
      long long a;
      cin >> a;
      if (a == 1) {
        long long pos;
        char c;
        cin >> pos >> c;
        alpha[int(s[pos - 1]) - 'a'].erase(pos);
        s[pos - 1] = c;
        alpha[int(c) - 'a'].insert(pos);
      } else {
        long long l, r;
        cin >> l >> r;
        long long ans = 0;
        for (long long i = 0; i < 26; i++) {
          if (*alpha[i].lower_bound(l) >= l && *alpha[i].lower_bound(l) <= r &&
              *alpha[i].lower_bound(l) > 0)
            ans++;
        }
        cout << ans << endl;
      }
    }
  }
  return 0;
}

```