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;
typedef long long ll;
vector< set<int> >q(30);
int main(){
    int n,i,j,t,l,r,k,f,ans;
    string s;
    char c;
    cin>>n>>s;
    for(int i=0;i<26;i++)q[i]={n};
    for(int i=0;i<n;i++)q[s[i]-'a'].insert(i);
    cin>>t;
    while(t--){
        cin>>f;
        if(f==1){
            cin>>k>>c;k--;
            if(s[k]!=c) q[s[k]-'a'].erase(k);
            s[k]=c;
            q[c-'a'].insert(k);
        }else{
            ans=0;
            cin>>l>>r;l--;r--;
            for(int j=0;j<26;j++)if(*q[j].lower_bound(l)<=r) ans++;
            cout<<ans<<endl;
        }
    }
    return 0;
}