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;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,Q; string S; cin >> N >> S >> Q;
    vector<set<int>> pos(26);
    for (int i=0;i<N;++i)
        pos[S[i]-'a'].insert(i);
    for (;Q--;){
        int t; cin >> t;
        if (t==1){
            int k; char c; cin >> k >> c; --k;
            for (int i=0;i<26;++i)
                if (pos[i].count(k)) pos[i].erase(k);
            pos[c-'a'].insert(k);
        } else {
            int l,r,ans=0; cin >> l >> r; --l,--r;
            for (int i=0;i<26;++i)
                if (pos[i].upper_bound(r)!=pos[i].lower_bound(l))
                    ++ans;
            cout << ans << '\n';
        }
    }
}