#include <bits/stdc++.h>
#define ll long long
#define mod 1000000007
#define upperlimit 1000100
#define INF 1000000100
#define INFL 1000000000000000100
#define eps 1e-8
#define endl '\n'
#define sd(n) scanf("%d",&n)
#define slld(n) scanf("%lld",&n)
#define pd(n) printf("%d",n)
#define plld(n) printf("%lld",n)
#define pds(n) printf("%d ",n)
#define pllds(n) printf("%lld ",n)
#define pdn(n) printf("%d\n",n)
#define plldn(n) printf("%lld\n",n)
#define REP(i,a,b) for(i=a;i<=b;i++)
#define mp make_pair
#define pb push_back
#define pcc pair<char,char>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vi vector<int>
#define vl vector<ll>
#define vii vector<pii>
#define vll vector<pll>
#define tr(container,it) for(typeof(container.begin()) it=container.begin();it!=container.end();it++)
#define F first
#define S second

using namespace std;

ll gcd(ll n1,ll n2){
	if(!n1)return n2;
	if(!n2)return n1;
	if(n1%n2==0)return n2;
	return gcd(n2,n1%n2);
}
ll powmod(ll base,ll exponent)
{
	base%=mod;
	ll ans=1;
	while(exponent){
		if(exponent&1)ans=(ans*base)%mod;
		base=(base*base)%mod;
		exponent/=2;
	}
	ans%=mod;
	return ans;
}
int cnt[26];
int cnt2[26][upperlimit+1];
int main()
{
	int i,j,k,n,q,Q,l,r,t;
	sd(n);
	sd(q);
	string s;
	cin>>s;
	for(i=0;i<n;i++){
		int temp=s[i]-'a';
        for(j=0;j<26;j++)cnt2[j][i+1]=cnt2[j][i];
        cnt2[temp][i+1]++;
		cnt[temp]++;
	}
    bool f=false;
	while(q--){
		sd(Q);
		sd(l);
		sd(r);
		if(Q==1){
            f=true;
			sd(t);
			for(i=l;i<=r;i++){
				int temp=s[i]-'a';
				s[i]=('a'+((temp+t)%26));
			}
		}
		else{
            ll answer=1;
            if(f){
                for(i=0;i<26;i++)cnt[i]=0;
                for(i=l;i<=r;i++){
                    cnt[s[i]-'a']++;
                }
                k=0;
                for(i=0;i<26;i++)if(cnt[i]){
                    k++;
                    answer=(answer*powmod(2,cnt[i]-1))%mod;
                }
                answer=((k+1)*answer-1)%mod;
            }
            else{
                k=0;
                for(i=0;i<26;i++)if(cnt2[i][r+1]>cnt2[i][l])k++;
                answer=((k+1)*powmod(2,r-l+1-k)-1)%mod;
            }
			plldn(answer);
		}
	}
	return 0;
}