// https://www.hackerrank.com/contests/101hack44/challenges/palindromic-subsets
/* author: github.com/bokonist */
#include "bits/stdc++.h"
using namespace std;
/* useful utility macros */
#define MIN(a,b,res) ((a)>(b))? (res)=(b):(res)=(a)
#define MAX(a,b,res) ((a)<(b))? (res)=(b):(res)=(a)
#define printArray(a,l,n) for(int(i)=(l);(i)<int(n);i++) { cout<<(a[i])<<' ';}
#define printWholeArray(a) for(int(i)=0;(i)<int((sizeof(a)/sizeof(a[0])));i++) { cout<<(a[i])<<' ';}
#define repeat(n) for(int (i)=0;(i)<(int)(n);++(i)) //repeat n times
#define repeatBounds(l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i)) //repeat lower to upperbound
#define until(x,val) for(int(i) = int(x); int(x)!=int(val); (i<val)?x++:x--) //until x becomes val through increment or decrement

/*some constants*/
static const long double PI = 3.142857143;

/*some shorthands*/
typedef long long ll;
#define op cout<< // output to stdout
#define ip cin>> //input from stdin
#define err cerr<< //output to stderr
#define nl cout<<"\n"; //newline
#define NO_OF_CHARS 256

bool canFormPalindrome(string str)
{
	if(str.length()==0)
		return false;
    int count[NO_OF_CHARS] = {0};
    for (int i = 0; str[i];  i++)
        count[str[i]]++;
    int odd = 0;
    for (int i = 0; i < NO_OF_CHARS; i++)
        if (count[i] & 1)
            odd++;
    return (odd <= 1);
}
ll generateSubsets(string &str)
{
	ll count=0;
	int len=str.length();
	string sub;
	for(ll i=0;i< pow(2,len);i++)
	{
		for(int j=0;j<len;j++)
		{
			if(i&(1<<j))
			{
				sub=sub+str[j];
			}
		}
		if(canFormPalindrome(sub))
			count++;
		sub="";
	}
	return count;
}
void shift(string &str, int i ,int j, int t)
{
	t=t%26;
	for (int index = i; index <= j; ++index)
	{
		str[index]+=t;
		if(str[index]>'z')
			str[index]-='a';
	}
}
int main()
{
	int n;
	ip n;
	int q;
	ip q;
	string str;
	ip str;
	while(q--)
	{
		int type;
		ip type;
		if(type==1)
		{
			int i,j,t;
			ip i>>j>>t;
			shift(str,i,j,t);
		}
		else
		{
			int i,j;
			ip i>>j;
			string sub= str.substr(i,j-i+1);
			op generateSubsets(sub);
			nl
		}
	}
	return 0;
}