#include<fstream>
#define NMAX 100000
using namespace std;
ifstream f("rmq.in"); ofstream g("rmq.out");
int n, t, x, y, log[NMAX];
int D[20][NMAX]; // 2^20 este aproximativ 100000
int main()
{
    f>>n>>t;
    log[1] = 0;
    for(int i = 2; i <= n; ++i) log[i] = log[i / 2] + 1;
    for(int i = 1; i <= n; ++i) f>>D[0][i];
    for(int i = 1; i <= log[n]; ++i)
        for(int j = 1; j <= n - (1<<(i - 1)); ++j)
            D[i][j] = min(D[i - 1][j], D[i - 1][j + (1<<(i - 1))]);
    while(t--)
    {
        f>>x>>y;
        int k = log[y - x + 1];
        g<<min(D[k][x], D[k][y - (1<<k) + 1])<<'\n';
    }
    g.close();
    return 0;
}