#include <fstream>

using namespace std;

ifstream fin("cuburi2.in");
ofstream fout("cuburi2.out");

typedef long long i64;

const int nmax= 250000;

i64 a[nmax+1], b[nmax+1], c[nmax+1];

string buffer;
string::iterator buffer_it;

void read_int_nn( int &x ) {
    for ( ; *buffer_it>'9' || *buffer_it<'0'; ++buffer_it ) ;
    for ( x= 0; *buffer_it<='9' && *buffer_it>='0'; ++buffer_it ) {
        x= x*10+*buffer_it-'0';
    }
}

int main(  ) {
    getline( fin, buffer, (char)0 );
    buffer_it= buffer.begin();

    int n, m;
    read_int_nn(n), read_int_nn(m);
    for ( int i= 1, x; i<=n; ++i ) {
        read_int_nn(x);
        a[i]= (i64)a[i-1]+x;
        b[i]= (i64)b[i-1]+a[i-1];
    }
    for ( int i= n-1; i>=1; --i ) {
        c[i]= (i64)c[i+1]+a[n]-a[i];
    }

    for ( int cnt= 1; cnt<=m; ++cnt ) {
        int x, y;
        read_int_nn(x), read_int_nn(y);

        i64 sol= x, step;
        for ( step= 1; step<=y-x; step*= 2 ) ;
        for ( ; step>0; step/= 2 ) {
            if ( (i64)sol+step<=y && (i64)a[sol+step-1]-a[x-1]<(i64)a[y]-a[sol+step-1] ) {
                sol+= step;
            }
        }

        i64 a1= (i64)b[sol]-b[x]-(sol-x)*a[x-1], a2= (i64)c[sol]-c[y]-(y-sol)*(a[n]-a[y]);
        fout<<sol<<" "<<(i64)a1+a2<<"\n";
    }

    return 0;
}