//SMALL TEST CASE

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
#define NMAX 100000
#define MOD 1000000007

char str[NMAX + 10];

int bit[26][NMAX + 10];
int N;

void update(int i, int ch)
{
    while(i <= N)
    {
        bit[ch][i]++;
        i += (i & -i);
    }
}

int count(int ch, int L, int R)
{
    int cnt2 = 0, cnt1 = 0;
    int ix = R;
    while(ix > 0)
    {
        cnt1 += bit[ch][ix];
        ix -= (ix & -ix);
    }
    
    ix = L - 1;
    while(ix > 0)
    {
        cnt2 += bit[ch][ix];
        ix -= (ix & -ix);
    }
    
    return cnt1 - cnt2;
}

long long int fast_pow(long long int base, long long int pow)
{
    if(pow == 0) return 1;
    if(pow == 1) return base;
    long long int temp = fast_pow(base, pow/2);
    if(pow % 2 == 0) return (temp %MOD * temp %MOD)%MOD;
    else return (temp %MOD * temp %MOD * base % MOD)%MOD;
}
int main() 
{
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int query;
    scanf("%d %d\n",&N,&query);
    scanf("%s",str + 1);
    str[N + 1] = '\0';
    for(int i = 1 ; i <= N; i++)
           update(i, str[i] - 'a'); 
    
    for(int t = 0; t < query; t++)
    {
        int type;
        scanf("%d",&type);
        //cout << type << "---"<< endl;
        if(type == 2)
        {
            int L, R;
            long long int ans = 0;
            scanf("%d %d",&L, &R);
            L++; R++;
            int ct[26], dist_letters = 0;
            memset(ct, 0 , sizeof(ct));
            for(int i = 0; i < 26; i++)
            {
                ct[i] = count(i, L, R);
                //printf("%d**",ct[i]);
                if(ct[i] > 0) dist_letters++;
            }
            
            if(R - L + 1 >= 3)
            {
                ans += (R - L + 1 - 2) * dist_letters * (dist_letters - 1);  
            }
            //cout << "***--" << ans << endl;
            for(int i = 0;  i < 26; i++)
            {
                 if(ct[i] == 0) continue;
                 ans += (fast_pow(2, ct[i]) % MOD - 1)%MOD;
            }
            printf("%lld\n",ans);
        }
        else
        {
            int L,R,U;
            cin >> L >> R >> U;
        }
    }
    return 0;
}