#include <bits/stdc++.h>

using namespace std;

const int SIZE = 10000;
struct InParsare
{
    char buffer[SIZE];
    int index;
    char nchar()
    {
        if(index == SIZE)
        {
            fread(buffer,1,SIZE,stdin);
            index = 0;
            return nchar();
        }
        return buffer[index++];
    }
    InParsare()
    {
        freopen("rmq.in","r",stdin);
        index = 0;
        fread(buffer,1,SIZE,stdin);
    }
    InParsare& operator >> (int &x)
    {
        x = 0;
        char c;
        for(c = nchar(); c < '0' || c > '9'; c = nchar());
        for(;c >= '0' && c <= '9'; c = nchar())
            x = x * 10 + c - '0';
        return *this;
    }
    void reload()
    {
        fseek(stdin,0,SEEK_SET);
        index = 0;
        fread(buffer,1,SIZE,stdin);
    }
}in;

const int NMAX = 100003;
int a[NMAX], n, m;
int dp[20][NMAX];

void rmq()
{
    for(int i=1;i<=n;i++)
        dp[0][i] = a[i];
    for(int i=1;(1<<i) <= n; i++)
    {
        for(int j=1;j<=n-(1<<i)+1;j++)
        {
            int power = 1<<(i-1);
            dp[i][j] = min(dp[i-1][j],dp[i-1][j+power]);
        }
    }
}

int main()
{
    freopen("rmq.out","w",stdout);
    in>>n>>m;
    for(int i=1;i<=n;i++)
        in>>a[i];
    rmq();
    for(int i=1;i<=m;i++)
    {
        int x,y;
        in>>x>>y;
        printf("%d\n",min(dp[((int)log2(y-x+1))][x],dp[((int)log2(y-x+1))][x+((y-x+1) - (1<<((int)log2(y-x+1))))]));
    }
    return 0;
}