You are given a string S of length N consisting of lowercase English letters.

Process Q queries of the following two types:

* Type 1: change the i_q-th character of S to c_q. (Do nothing if the i_q-th character is already c_q.)
* Type 2: answer the number of different characters occurring in the substring of S between the l_q-th and r_q-th characters (inclusive).

Constraints

* N, Q, i_q, l_q, and r_q are integers.
* S is a string consisting of lowercase English letters.
* c_q is a lowercase English letter.
* 1 \leq N \leq 500000
* 1 \leq Q \leq 20000
* |S| = N
* 1 \leq i_q \leq N
* 1 \leq l_q \leq r_q \leq N
* There is at least one query of type 2 in each testcase.

Input

Input is given from Standard Input in the following format:


N
S
Q
Query_1
\vdots
Query_Q


Here, Query_i in the 4-th through (Q+3)-th lines is one of the following:


1 i_q c_q



2 l_q r_q


Output

For each query of type 2, print a line containing the answer.

Example

Input

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7


Output

3
1
5
#include <bits/stdc++.h>
using namespace std;

vector < set<int> > adj(28);
string s;
char c;
int q, a, b, d, ans;
int main()
{
 	int n;
    cin >> n >> s >> q;
    s = 'a' + s;
    for(int i = 1; i < s.size(); ++i)
        adj[s[i] - 'a'].insert(i);
    set<int>::iterator it;
    while(q--)
    {
        cin >> a;
        if(a == 1)
        {
            cin >> b >> c;
            adj[s[b] - 'a'].erase(b);
            s[b] = c;
            adj[c - 'a'].insert(b);
        }
        else
        {
            ans = 0;
            cin >> b >> d;
            for(int i = 0; i < 27; ++i)
            {
                it = adj[i].lower_bound(b);
              	
                ans += (*it <= d && *it >= b);
            }
            cout << ans << '\n';
        }
    }
}