https://ac.nowcoder.com/acm/contest/78292/F

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PII;
typedef vector<long long> VI;
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
#define pb(i) push_back(i)
#define int long long
#define INF 0x3f3f3f3f
#define oz  998244353
#define endl '\n'
#define N 200010
const int mod = 1e9 + 7;


int p[N], si[N];
int find(int x) {
    if (x != p[x]) p[x] = find(p[x]);
    return p[x];
}
//size[find(b)] += size[find(a)];
//p[find(a)] = find(b);


//无论多长的回文串，一定包含 1 2 1 或 1 1 1
//则只要 la[i] = i - 2则一定存在回文串

void solve() {
    int n, q;
    cin >> n >> q;

    VI a(n + 1);
    rep(i, 1, n)cin >> a[i];

    VI la(n + 1); // la[i]表示上一次出现a[i]的位置;
    map<int, int> mp;

    rep(i, 1, n) {
        la[i] = mp[a[i]];
        mp[a[i]] = i;
    }

    per(i, n, 1) {
        if (la[i] == i - 1) {
            la[i] = la[la[i]];
        }
    }

    VI maxla(n); //维护区间最大值 因为l之前不可能有比l大的数 所以maxla[r] >= l 则该数一定出现在 l 到 r之间

    rep(i, 1, n) {
        maxla[i] = max(maxla[i - 1], la[i]);
    }

    while (q --) {
        int l, r;
        cin >> l >> r;

        if (maxla[r] >= l)cout << "YES" << endl;
        else cout << "NO" << endl;
    }


}

signed main() {

    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
//  cin >> T;
    while (T --)

        solve();

    return 0;
}