#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, k, q;
cin >> n >> k >> q;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++) cin >> a[i];
vector<int> dp(n + 1, n);
// For every adjacency i -> i+1, compress the unordered pair.
// x[i], y[i] with x[i] < y[i]
// t[i] = 0 if a[i] < a[i+1], 1 otherwise
// badself if a[i] == a[i+1]
vector<int> x(n), y(n), t(n), self(n);
for (int i = 1; i < n; i++) {
if (a[i] == a[i + 1]) {
self[i] = 1;
x[i] = y[i] = a[i];
t[i] = 0;
} else {
self[i] = 0;
x[i] = min(a[i], a[i + 1]);
y[i] = max(a[i], a[i + 1]);
t[i] = (a[i] > a[i + 1]);
}
}
// For current left endpoint l, an edge at position i means:
// if (i-l)%2==0 then required relation is a[i] > a[i+1]
// else required relation is a[i] < a[i+1]
//
// Suppose edge i is between u and v, with u < v.
// If t[i]=0, then actual order in array is u,v.
// If t[i]=1, then actual order in array is v,u.
//
// We maintain whether pair (u,v) is forced as u<v or u>v in current window.
// dir = t[i] xor ((i-l)%2==0 ? 1 : 0)
// since when parity says ">", if array order is u,v then requirement is u>v.
//
// To support sliding l, we process separately for l parity 1 and 0.
// For fixed l parity, (i-l)%2 depends only on parity of i.
auto solve_parity = [&](int start_parity, vector<int>& ans) {
unordered_map<long long, int> cnt0, cnt1;
cnt0.reserve(n * 2);
cnt1.reserve(n * 2);
cnt0.max_load_factor(0.7f);
cnt1.max_load_factor(0.7f);
auto key = [&](int u, int v) -> long long {
return 1LL * u * (k + 1) + v;
};
int bad = 0;
int eq = 0;
int r = start_parity; // window over edges [l..r-1], so r is current right endpoint in array positions
if (r == 0) r = 2;
for (int l = start_parity; l <= n; l += 2) {
if (r < l) r = l;
while (r < n) {
int i = r; // adding edge i = (r, r+1)
int add_eq = self[i];
long long kk = key(x[i], y[i]);
int dir;
if ((i & 1) == (l & 1)) dir = t[i] ^ 1;
else dir = t[i];
int old0 = 0, old1 = 0;
if (!self[i]) {
auto it0 = cnt0.find(kk);
if (it0 != cnt0.end()) old0 = it0->second;
auto it1 = cnt1.find(kk);
if (it1 != cnt1.end()) old1 = it1->second;
}
int nbad = bad;
int neq = eq + add_eq;
if (!self[i]) {
if (dir == 0) {
if (old0 == 0 && old1 > 0) nbad++;
} else {
if (old1 == 0 && old0 > 0) nbad++;
}
}
if (neq > 0 || nbad > 0) break;
eq = neq;
bad = nbad;
if (!self[i]) {
if (dir == 0) cnt0[kk]++;
else cnt1[kk]++;
}
r++;
}
ans[l] = r;
if (l < n) {
int i = l; // remove edge l
if (self[i]) {
eq--;
} else {
long long kk = key(x[i], y[i]);
int dir;
if ((i & 1) == (l & 1)) dir = t[i] ^ 1;
else dir = t[i];
int old0 = cnt0[kk];
int old1 = cnt1[kk];
if (dir == 0) {
if (old0 == 1 && old1 > 0) bad--;
cnt0[kk]--;
if (cnt0[kk] == 0) cnt0.erase(kk);
} else {
if (old1 == 1 && old0 > 0) bad--;
cnt1[kk]--;
if (cnt1[kk] == 0) cnt1.erase(kk);
}
}
}
}
};
solve_parity(1, dp);
if (n >= 2) solve_parity(2, dp);
while (q--) {
int l, r;
cin >> l >> r;
cout << (dp[l] >= r ? "YES\n" : "NO\n");
}
return 0;
}
