Problem:
Score : 600 points

Problem Statement

We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.

You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?

Constraints

1\leq N,Q \leq 5 \times 10^5

1\leq c_i \leq N

1\leq l_i \leq r_i \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q

Output

Print Q lines. The i-th line should contain the response to the i-th query.

Sample Input 1

4 3
1 2 1 3
1 3
2 4
3 3

Sample Output 1

2
3
1

The 1-st, 2-nd, and 3-rd balls from the left have the colors 1, 2, and 1 - two different colors.

The 2-st, 3-rd, and 4-th balls from the left have the colors 2, 1, and 3 - three different colors.

The 3-rd ball from the left has the color 1 - just one color.

Sample Input 2

10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8

Sample Output 2

1
2
2
1
2
2
6
3
3
3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)


#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#include <prettyprint.hpp>
#define debug(...)  cerr << "[" << #__VA_ARGS__ << "]: ", d_err(__VA_ARGS__);
#else
#define debug(...) 83;
#endif

void d_err() {
    cerr << endl;
}

template <typename H, typename... T>
void d_err(H h, T... t) {
    cerr << h << " ";
    d_err(t...);
}

template <typename T>
void print(T x) {
    cout << x << endl;
}

#define ALL(x) (x).begin(), (x).end()
#define FOR(i, m, n) for (int i = (m); i < (n); ++i)
#define REVFOR(i, m, n) for (int i = (n - 1); i >= (m); --i)
#define REP(i, n) FOR(i, 0, n)
#define REVREP(i, n) REVFOR(i, 0, n)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define bcnt __builtin_popcountll

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<ll,ll> Pll;
typedef pair<int,int> Pin;

ll INF = 1e16;
int inf = 1e9;
ll MOD = 1e9+7;

template<typename T>
struct BIT {
    int N;
    vector<T> dat;

    // 0-indexing
    explicit BIT(int n): N(n), dat(N + 1, 0) {}

    T sum(int i) {
        T s = T();
        for(++i; i > 0; i -= i & -i) s += dat[i];
        return s;
    }

    void add(int i, const T& x) {
        for (++i; i <= N; i += i & -i) dat[i] += x;
    }

    // min_x {x | v1 + v2 + ⋯ + vx ≧ w}
    // Assuming v_q >= 0
    int lower_bound(T w) {
        if (w <= 0) return 0;

        int x = 0; T k = 1;
        while (k * 2 <= N) k <<= 1;
        for (; k > 0; k /= 2) {
            if (x + k <= N && dat[x + k] < w) {
                w -= dat[x + k];
                x += k;
            }
        }
        return x + 1;
    }
};

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    
    int N, Q;
    cin >> N >> Q;

    vll c(N);
    REP(i, N) cin >> c[i], c[i]--;

    vector<vector<pair<int, int>>> query(N);
    REP(i, Q) {
        int l, r; cin >> l >> r;
        --l; --r;
        query[r].eb(l, i);
    }

    vector<int> last(N, -1);
    BIT<int> bit(N);
    vector<int> ans(Q);
    REP(i, N) {
        if (~last[c[i]]) bit.add(last[c[i]], -1);
        last[c[i]] = i;
        bit.add(i, 1);

        for(auto &p: query[i]) {
            ans[p.se] = bit.sum(i) - bit.sum(p.fi-1);
        }
    }

    for (auto &e: ans) {
        print(e);
    }
    
}