//回滾莫隊 裸題
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int,int>;
#define fs first
#define sc second
int n, q, a[510000], cnt[510000] = {0};
long long nowans, ans[510000];
vector <int> b;
int B;
vector <pair<pii,int>> qs[510000];
void add (int x) {
cnt[x]++;
nowans = max(nowans, b[x] * 1LL * cnt[x]);
}
int main () {
ios_base::sync_with_stdio(false); cin.tie(0);
cin >> n >> q;
B = sqrt((n + q + 1) / 2);
for (int i = 1; i <= n; i++) {
cin >> a[i];
b.push_back(a[i]);
}
sort (b.begin(), b.end());
b.erase(unique(b.begin(), b.end()), b.end());
for (int i = 1; i <= n; i++) a[i] = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
for (int i = 1; i <= q; i++) {
pii p;
cin >> p.fs >> p.sc;
int lb = p.fs / B, rb = p.sc / B;
if (lb == rb) { // 特判塊太小
for (int j = p.fs; j <= p.sc; j++) cnt[a[j]] = 0;
nowans = 0;
for (int j = p.fs; j <= p.sc; j++) add(a[j]);
ans[i] = nowans;
continue;
}
qs[p.fs / B] . push_back({p, i});
}
for (int i = 1 / B; i <= n / B; i++) {
sort(qs[i].begin(), qs[i].end(), [](pair<pii,int>a, pair<pii,int>b){return a.fs.sc < b.fs.sc;} );
int RB = min(i * B + B, n), R = RB - 1;
nowans = 0;
memset(cnt, 0, sizeof(cnt));
for (auto qq : qs[i]) {
while (R < qq.fs.sc) {
R++;
add(a[R]);
}
long long oldans = nowans;
vector <pii> old;
for (int j = RB - 1; j >= qq.fs.fs; j--) old.push_back({a[j], cnt[a[j]]});
for (int j = RB - 1; j >= qq.fs.fs; j--) add(a[j]);
ans[qq.sc] = nowans;
nowans = oldans;
for (pii p : old) cnt[p.fs] = p.sc;
}
}
for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
return 0;}
