#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define ar array
#define vt vector
#define pq priority_queue
#define pu push
#define pub push_back
#define em emplace
#define emb emplace_back

#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()
#define allp(x, l, r) x.begin() + l, x.begin() + r
#define len(x) (int)x.size()

using namespace std;
using namespace __gnu_pbds;

using ll = long long;
using ld = long double;
using ull = unsigned long long;

template <class T, size_t N>
void re(array <T, N>& x);
template <class T> 
void re(vt <T>& x);

template <class T> 
void re(T& x) {
    cin >> x;
}

template <class T, class... M> 
void re(T& x, M&... args) {
    re(x); re(args...);
}

template <class T> 
void re(vt <T>& x) {
    for(auto& it : x)
        re(it);
}

template <class T, size_t N>
void re(array <T, N>& x) {
    for(auto& it : x)
        re(it);
}

template <class T, size_t N>
void wr(array <T, N> x);
template <class T> 
void wr(vt <T> x);

template <class T> 
void wr(T x) {
    cout << x;
}

template <class T, class ...M>  void wr(T x, M... args) {
    wr(x), wr(args...);
}

template <class T> 
void wr(vt <T> x) {
    for(auto it : x)
        wr(it, ' ');
}

template <class T, size_t N>
void wr(array <T, N> x) {
    for(auto it : x)
        wr(it, ' ');
}


inline void Open(const string Name) {
    #ifndef ONLINE_JUDGE
        (void)!freopen((Name + ".in").c_str(), "r", stdin);
        (void)!freopen((Name + ".out").c_str(), "w", stdout);
    #endif
}

void solve() {
    const int INF = 1e9;
    const int K = 17;
    int n, q; re(n, q);
    vt <int> v(n); re(v);

    vt <vt <int>> st(K, vt <int>(n));
    /* To precompute the answer and answer the queries in constant time we will use a sparse table:
       We divide an interval n into different intervals whose lengths are powers of 2. To better understand let's take the following example:

       For the array [1, 5, 6, 4, 3] we will store the values in the spare table as such:
       i = 0 -> [1, 5, 6, 4, 3]
       i = 1 -> [1, 5, 6, 4, 3] -> [min{1, 5}, min{5, 6}, min{6, 4}, min{4, 3}] -> [1, 5, 4, 3]
       i = 2 -> [1, 5, 4, 3] -> [min{1, 4}, min{5, 3}] -> [1, 3]

    */

    st[0] = v;
    for (int i = 1; i < K; ++i)
        for (int j = 0; j + (1 << (i - 1)) < n; ++j) {  
            st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
        }

    for (int i = 0; i < q; ++i) {
        int l, r; re(l, r);
        int ans = INF;

        --l, --r;
        /* iterative better for sum queries
        for (int j = K - 1; j >= 0; --j)
            if (l + (1 << j) - 1 <= r) {
                ans = min(ans, st[j][l]);
                l += (1 << j);
            }
        */

        /* optimized better for overlapping functions such as gcd, min, max, ... */
        int k = 31 - __builtin_clz(r - l + 1);
        ans = min(st[k][l], st[k][r - (1 << k) + 1]);
        wr(ans, '\n');
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    Open("rmq");

    int t = 1;
    for(;t;t--) {
        solve();
    }
    
    return 0;
}