#include<bits/stdc++.h>
using namespace std;

#define rep(i, l, r) for (int i = (int)(l); i<(int)(r); i++)
#define ll long long
#define all(x) (x).begin(), (x).end()
#define siz(x) (int)(x).size()

const int inf = 1e9;
const ll INF = 4e18;

template<class T>bool chmax(T &a, T b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, T b) { if (b<a) { a=b; return 1; } return 0; }

vector<int> di = {-1, 0, 1, 0}, dj = {0, -1, 0, 1};

//1次元累積和
//全て0-indexedで処理
template<class T>
struct CumulativeSum {
    int siz;
    vector<T> S;
    bool done;
    CumulativeSum() : CumulativeSum(0) {}
    CumulativeSum(int N) : CumulativeSum(vector<T>(N, 0)) {}
    //累積和の構築はしない
    CumulativeSum(vector<T> A) {
        done = false;
        siz = (int)A.size();
        S.resize(siz+1);
        S[0] = 0;
        for (int i = 0; i < siz; i++) {
            S[i+1] = A[i];
        }
    }
    //累積
    void build() {
        assert(!done);
        for (int i = 1; i <= siz; i++) {
            S[i] += S[i-1];
        }
        done = true;
    }
    //加算（累積前のみ）
    T add(int idx, T a) {
        assert(!done);
        return S[idx+1] += a;
    }
    //代入（累積前のみ）
    T set(int idx, T a) {
        assert(!done);
        return S[idx+1] = a;
    }
    //取得（累積前、累積後両方とも可能だが、どちらを想定するかをexpected_doneで渡す）
    [[nodiscard]]
    T get(int idx, bool expected_done) {
        assert(expected_done == done);
        return S[idx+1];
    }
    //区間和取得（累積後のみ）
    //半開区間で与える
    [[nodiscard]]
    T sum(int l, int r) {
        assert(done);
        return S[r]-S[l];
    }
};

void solve() {
    int N, Q; cin >> N >> Q;
    vector<int> A(N); rep(i, 0, N) cin >> A[i];
    vector<int> L(Q), R(Q);
    rep(i, 0, Q) {
        cin >> L[i] >> R[i];
        L[i]--;
    }
    vector<ll> ans(Q, 0);
    rep(b, 0, 26) {
        vector<int> B(N);
        rep(i, 0, N) B[i] = (A[i]>>b&1);
        // rep(i, 0, N) cout << B[i] << " ";
        // cout << endl;
        CumulativeSum<int> C(B);
        C.build();
        rep(i, 0, Q) {
            int cnt = C.sum(L[i], R[i]);
            // cout << cnt << endl;
            ll x = (ll)cnt * (R[i]-L[i]-cnt);
            ans[i] += x * (1LL<<b);
        }
    }
    rep(i, 0, Q) cout << ans[i] << endl;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int T = 1; 
    // cin >> T;
    while(T--) {
        solve();
    }
}