#include <fstream>

using namespace std;

const int NMAX = 1e6 + 1, KMAX = 17;

class InParser
{
private:
    FILE *fin;
    char *buff;
    int sp;

    char read_ch()
    {
        ++sp;
        if (sp == 4096)
        {
            sp = 0;
            fread(buff, 1, 4096, fin);
        }
        return buff[sp];
    }

public:
    InParser(const char* nume)
    {
        fin = fopen(nume, "r");
        buff = new char[4096]();
        sp = 4095;
    }

    InParser& operator >> (int &n)
    {
        char c;
        while (!isdigit(c = read_ch()) && c != '-');
        int sgn = 1;
        n = c - '0';
        while (isdigit(c = read_ch())) n = 10 * n + c - '0';
        n *= sgn;
        return *this;
    }
};

class OutParser
{
private:
    FILE *fout;
    char *buff;
    int sp;

    void write_ch(char ch)
    {
        if (sp == 50000)
        {
            fwrite(buff, 1, 50000, fout);
            sp = 0;
            buff[sp++] = ch;
        }
        else
        {
            buff[sp++] = ch;
        }
    }


public:
    OutParser(const char* name)
    {
        fout = fopen(name, "w");
        buff = new char[50000]();
        sp = 0;
    }
    ~OutParser()
    {
        fwrite(buff, 1, sp, fout);
        fclose(fout);
    }

    OutParser& operator << (int vu32)
    {
        if (vu32 <= 9)
        {
            write_ch(vu32 + '0');
        }
        else
        {
            (*this) << (vu32 / 10);
            write_ch(vu32 % 10 + '0');
        }
        return *this;
    }

    OutParser& operator << (char ch)
    {
        write_ch(ch);
        return *this;
    }
};

InParser fin("rmq.in");
OutParser fout("rmq.out");

int n, m, lg2[NMAX], t[KMAX][NMAX], x, y, k;

int main()
{
    fin >> n >> m;
    lg2[0] = -1; ///(*)
    for(int i = 1; i <= n; ++i)
    {
        fin >> t[0][i];
        lg2[i] = lg2[i >> 1] + 1;
    }
    lg2[0] = 0;
    for(int i = 1; i <= lg2[n]; ++i)
        for(int j = 1; j + (1 << i) - 1 <= n; ++j)
            t[i][j] = min(t[i - 1][j], t[i - 1][j + (1 << (i - 1))]);
    while(m--)
    {
        fin >> x >> y;
        k = lg2[y - x + 1];
        fout << min(t[k][x], t[k][y - (1 << k) + 1]) << '\n';
    }
    return 0;
}