#include <bits/stdc++.h>
#include <cassert>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll = long long;
using mint = modint998244353;
//using mint = modint1000000007;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repu(i, s, t) for (int i = (int)(s); i < (int)(t); i++)
#define repd(i, s, t) for (int i = (int)(s)-1; i >= (int)(t); i--)
#define all(v) v.begin(), v.end()
template<typename T> bool chmax(T &a, const T b) { if(a >= b) return false; a = b; return true; }
template<typename T> bool chmin(T &a, const T b) { if(a <= b) return false; a = b; return true; }
template<typename T> istream& operator>>(istream &in, vector<T> &a) { for(T &x: a) in >> x; return in; }
template<typename T> ostream& operator<<(ostream &out, const vector<T> &a) { for(const T &x: a) out << x << ' '; return out; }
const int di[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dj[] = {0, 1, 0, -1, -1, 1, 1, -1, 0};


struct S {
    bool zero_ex, one_ex;
    vector<vector<mint>> val;
    S() : zero_ex(false), one_ex(false), val(2, vector<mint>(3)) {}
    S(bool x) : zero_ex(!x), one_ex(x), val(2, vector<mint>(3)) {
        val[int(x)] = {1, 1, 1};
    }
};
S op(S l, S r) {
    S res;
    rep(i, 2) rep(j, 3) res.val[i][j] = l.val[i][j] + l.val[i][0] * r.val[0][j] + l.val[i][1] * r.val[1][j];
    if(!l.zero_ex) res.val[0] = r.val[0];
    if(!l.one_ex) res.val[1] = r.val[1];
    if(r.zero_ex) rep(i, 2) res.val[i][0] -= l.val[i][0];
    if(r.one_ex) rep(i, 2) res.val[i][1] -= l.val[i][1];
    res.zero_ex = l.zero_ex || r.zero_ex;
    res.one_ex = l.one_ex || r.one_ex;
    return res;
}
S e() { return S(); }
using F = bool;
S mapping(F l, S r) {
    if(l) return e();
    return r;
}
F composition(F l, F r) { return l || r; }
F id() { return false; }

int main() {
    int n, q;
    string s;
    cin >> n >> q >> s;

    lazy_segtree<S, op, e, F, mapping, composition, id> lst(n);
    rep(i, n) lst.set(i, S(s[i]-'0'));
    while(q--) {
        int t, l, r;
        cin >> t >> l >> r;
        l--;
        if(t == 1) {
            lst.apply(l, r, true);
        } else {
            cout << lst.prod(l, r).val[1][2].val() << '\n';
        }
    }
    return 0;
}