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<iostream>
#include<set>
using namespace std;

int main(){
    int n, q;
    string s;
    cin >> n >> s >> q;
    set<int> t[26];
    for(int i = 0; i < n; i++)  t[s[i]-'a'].insert(i+1);
    for(int i = 0; i < 26; i++) t[i].insert(n+1);
    while(q--){
        int op;
        cin >> op;
        if(op == 1){
            int i;
            char c;
            cin >> i >> c;
            i--;
            t[s[i]-'a'].erase(i+1);
            s[i] = c;
            t[s[i]-'a'].insert(i+1);
        }else if(op == 2){
            int l, r;
            cin >> l >> r;
            int ans = 0;
            for(int i = 0; i < 26; i++) ans += *(t[i].lower_bound(l)) <= r;
            cout << ans << endl;
        }
    }
    return 0;
}