#include <bits/stdc++.h>
#define ALL(a) (a).begin(), (a).end()
#define FOR(x,n) for(int x = 0; x < n; x++)
#define SZ(a) ((int)(a).size())
using namespace std;
typedef long long ll;
typedef vector<int> vi;

int N, M;
string s;

const ll MOD = 1e9+7LL;

const int MXN = 1e5 + 1;
int t[4*MXN][26] = {}, LAZY[4*MXN] = {}, tmpA[26] = {};

void build(int n = 1, int l = 0, int r = N-1) {
  if(l==r) {
    t[n][s[l]-'a'] = 1;
  } else {
    int mid = (l+r)/2;
    build(2*n,l,mid); build(2*n+1,mid+1,r);
    FOR(x,26) t[n][x] = t[2*n][x] + t[2*n+1][x];
  }
}

void push(int n, int l, int r) {
  if(LAZY[n] %= 26) {
    FOR(x,26) tmpA[x] = t[n][x];
    FOR(x,26) t[n][x] = tmpA[(x-LAZY[n]+26)%26];
    if(l != r) {
      LAZY[2*n] = (LAZY[2*n] + LAZY[n]) % 26;
      LAZY[2*n+1] = (LAZY[2*n+1] + LAZY[n]) % 26;
    }
    LAZY[n] = 0;
  }
}

void update(int a, int b, int m, int n=1, int l=0, int r=N-1) {
  push(n,l,r);
  if(l > b || r < a) return;
  else if(a <= l && r <= b) {
    LAZY[n] = (LAZY[n] + m)%26;
  } else {
    int mid = (l+r)/2;
    update(a,b,m,2*n,l,mid); update(a,b,m,2*n+1,mid+1,r);
    push(2*n,l,mid); push(2*n+1,mid+1,r);
    FOR(x,26) t[n][x] = t[2*n][x] + t[2*n+1][x];
  }
}

vi query(int a, int b, int n=1, int l=0, int r=N-1) {
  push(n,l,r);
  if(l > b || r < a) return vi(26);
  else if(a <= l && r <= b) {
    vi tmp = vector<int>(26);
    FOR(x,26) tmp[x] = t[n][x];
    return tmp;
  } else {
    int mid = (l+r)/2;
    vi tmp1 = query(a,b,2*n,l,mid);
    vi tmp2 = query(a,b,2*n+1,mid+1,r);
    FOR(x,26) tmp1[x] += tmp2[x];
    return tmp1;
  }
}

ll mm[MXN] = {}, mm2[MXN] = {}, facts[MXN], finv[MXN];

ll cmb(ll i, ll j) {
  return (((facts[i] * finv[j])%MOD) * finv[i-j])%MOD;
}

ll solve2(ll i) {
  if(mm[i] != -1) {return mm[i]; }
  mm[i] = 0;
  for(ll j = 0; j <= i; j+=2) {
    mm[i] = (mm[i] + cmb(i,j))%MOD;
  }
  return mm[i];
}

ll solve3(ll i) {
  if(mm2[i] != -1) { return mm2[i]; }
  mm2[i] = 0;
  for(ll j = 1; j <= i; j+=2)
    mm2[i] = (mm2[i] + cmb(i,j))%MOD;
  return mm2[i];
}

ll solve(int a, int b) {
  vi tmp = query(a,b);
  ll ans = 1, ones = 0;
  FOR(x,26) {
    ll tmp1 = solve3(tmp[x]);
    FOR(y,26) if(x != y) tmp1 = (tmp1 * solve2(tmp[y]))%MOD;
    ones = (ones + tmp1)%MOD;
  }
  FOR(x,26) ans = (ans * solve2(tmp[x]))%MOD;
  return (((ans-1+ones)%MOD)+MOD)%MOD;
}

ll fE(ll b, ll e) {
  ll r = 1;
  while(e) if(e&1) r=(r*b)%MOD,e--; else b=(b*b)%MOD, e/=2;
  return r;
}

int main() {
  finv[0] = facts[0] = 1;
  for(ll x = 1; x < MXN; x++)
    facts[x] = (facts[x-1] * x)%MOD, finv[x] = fE(facts[x],MOD-2);
  memset(mm,-1,sizeof mm);
  memset(mm2,-1,sizeof mm2);
  cin >> N >> M >> s;
  build();
  FOR(x,M) {
    int type; cin >> type;
    if(type == 2) {
      int a, b; cin >> a >> b;
      cout << solve(a,b) << "\n";
    } else {
      int a, b, m; cin >> a >> b >> m;
      update(a,b,m);
    }
  }
}