// https://www.infoarena.ro/problema/rmq

// sparse table pe minim, query O(1) dupa preprocesare O(n log n)
// min e idempotent => cele doua blocuri de lungime 2^k se pot suprapune

#include <fstream>
#include <algorithm>

using namespace std;

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

const int MAXN = 100001;
const int LOG = 17;

int a[MAXN];
int sp[LOG][MAXN];
int lg[MAXN];

int main() {
    int n, m;
    fin >> n >> m;
    for (int i = 0; i < n; i++)
        fin >> a[i];

    lg[1] = 0;
    for (int i = 2; i <= n; i++)
        lg[i] = lg[i / 2] + 1;

    for (int i = 0; i < n; i++)
        sp[0][i] = a[i];
    for (int k = 1; (1 << k) <= n; k++)
        for (int i = 0; i + (1 << k) <= n; i++)
            sp[k][i] = min(sp[k - 1][i], sp[k - 1][i + (1 << (k - 1))]);

    for (int q = 0; q < m; q++) {
        int x, y;
        fin >> x >> y;
        int l = x - 1, r = y - 1;   // problema indexeaza de la 1
        int k = lg[r - l + 1];
        fout << min(sp[k][l], sp[k][r - (1 << k) + 1]) << '\n';
    }
    return 0;
}