#include <algorithm>
#include <cassert>
#include <iostream>
#include <numeric>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;


#include <atcoder/modint>
using mint = atcoder::modint998244353;

// Rolling Hash (Rabin-Karp), 1dim
template <typename V> struct rolling_hash {
    int N;
    const V B;
    std::vector<V> hash;         // hash[i] = s[0] * B^(i - 1) + ... + s[i - 1]
    static std::vector<V> power; // power[i] = B^i
    void _extend_powvec() {
        if (power.size() > 1 and power.at(1) != B) power = {V(1)};
        while (static_cast<int>(power.size()) <= N) {
            auto tmp = power.back() * B;
            power.push_back(tmp);
        }
    }
    template <typename Int>
    rolling_hash(const std::vector<Int> &s, V b = V::randgen()) : N(s.size()), B(b), hash(N + 1) {
        for (int i = 0; i < N; i++) hash[i + 1] = hash[i] * B + s[i];
        _extend_powvec();
    }
    rolling_hash(const std::string &s = "", V b = V::randgen()) : N(s.size()), B(b), hash(N + 1) {
        for (int i = 0; i < N; i++) hash[i + 1] = hash[i] * B + s[i];
        _extend_powvec();
    }
    void addchar(const char &c) {
        V hnew = hash[N] * B + c;
        N++, hash.emplace_back(hnew);
        _extend_powvec();
    }
    V get(int l, int r) const { // s[l] * B^(r - l - 1) + ... + s[r - 1]
        return hash[r] - hash[l] * power[r - l];
    }
    int lcplen(int l1, int l2) const { return longest_common_prefix(*this, l1, *this, l2); }
};
template <typename V> std::vector<V> rolling_hash<V>::power{V(1)};


int main() {
    cin.tie(nullptr), ios::sync_with_stdio(false);

    int N, Q;
    cin >> N >> Q;
    vector<mint> A(N);
    for (auto &x : A) {
        int y;
        cin >> y;
        x = y;
    }

    const rolling_hash<mint> rh(A, mint(2).inv());

    vector<int> last(N);
    for (int l = 0, i = 1; i < N; ++i) {
        if (A.at(i - 1).val() >= A.at(i).val()) l = i;
        last.at(i) = l;
    }

    while (Q--) {
        int l, r;
        cin >> l >> r;
        --l;
        int ll = max(l, last.at(r - 1));
        int len = r - ll;
        cout << (A.at(ll) + (len >= 2 ? rh.get(ll + 1, r) * mint(2).pow(len - 2) : 0)).val() << '\n';
    }
}