#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

int pali(string s)
{
    map<string, int> m;
    int n = s.size();
 
    
    int R[2][n+1];
 
    
    s = "@" + s + "#";
 
    for (int j = 0; j <= 1; j++)
    {
        int rp = 0;   // length of 'palindrome radius'
        R[j][0] = 0;
 
        int i = 1;
        while (i <= n)
        {
    
            while (s[i - rp - 1] == s[i + j + rp])
                rp++;  // Incrementing the length of palindromic
                       // radius as and when we find vaid palindrome
 
    R[j][i] = rp;
            int k = 1;
            while ((R[j][i - k] != rp - k) && (k < rp))
            {
                R[j][i + k] = min(R[j][i - k],rp - k);
                k++;
            }
            rp = max(rp - k,0);
            i += k;
        }
    }
 
    
    s = s.substr(1, n);
 
    
    m[string(1, s[0])]=1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= 1; j++)
            for (int rp = R[j][i]; rp > 0; rp--)
               m[s.substr(i - rp - 1, 2 * rp + j)]=1;
        m[string(1, s[i])]=1;
    }
 
    //printing all distinct palindromes from hash map
   //cout << "Below are " << m.size()-1
      //  << " palindrome sub-strings";
   //map<string, int>::iterator ii;
   //for (ii = m.begin(); ii!=m.end(); ++ii)
      //cout << (*ii).first << endl;
    return m.size()-1;
}

int main(){
    int n;
    int q;
    cin >> n >> q;
    string s;
    cin >> s;
    int inp1=0;
    int start,end,t;
    int q1,q2;
    string ns;
    long long int ans;
    for(int a0 = 0; a0 < q; a0++){
        cin>>inp1;
        if(inp1==1)
            {
            cin>>start>>end>>t;
            for(int i=start;i<=end;i++)
                {
                q1=s[i]-'a';
                q2=(q1+t)%26;
                s[i]='a'+q2;
            }
           // cout<<s<<endl;
        }
        else
        {
            cin>>start>>end;
            ns = s.substr(start,end-start+1);
            ans = pali(ns)%(1000000007);
            cout<<ans<<endl;
        }
        
    }
   
    return 0;
}