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)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef vector<vector<int>> VVI;
typedef vector<ll> VLL;
typedef vector<vector<ll>> VVLL;
typedef pair<ll, ll> Pair;
template <class T> T input() {T t; cin>>t; return t;}
#define FOR(i,a,b) for (ll i = a; i < b; i++)
#define REP(i,b) FOR(i,0,b)
#define RFOR(i,a,b) for (ll i = a-1; i >= b; i--)
#define RREP(i,a) RFOR(i,a,0)
#define REPALL(i,x) for (ll i = 0; i < x.size(); i++)
#define RREPALL(i,x) for (ll i = x.size()-1; i >= 0; i--)
#define REPITR(itr,x) for (auto itr = (x).begin(); itr != (x).end(); itr++)
#define ALL(x) (x).begin(), (x).end()
#define SORT(x) sort(ALL(x))
#define MIN_ELEMENT(x) min_element(ALL(x))
#define MAX_ELEMENT(x) max_element(ALL(x))
#define COUNT(x,num) count(ALL(x), num)
#define MEMSET(x,val) memset(x, val, sizeof(x))
#define CHMAX(a,b) a = max(a,b)
#define CHMIN(a,b) a = min(a,b)
#define FIND(x,key) x.find(key) != x.end()
#define IN_RANGE_V2(v,k,x,y) (0<=x+dx[k]&&x+dx[k]<v[y].size()&&0<=y+dy[k]&&y+dy[k]<v.size())
#define debug(x) cerr<<__LINE__<<": "<<(#x)<<" = "<<(x)<<endl;
void YES(bool flag) {cout<<(flag ? "YES" : "NO")<<endl;}
void Yes(bool flag) {cout<<(flag ? "Yes" : "No")<<endl;}
void yes(bool flag) {cout<<(flag ? "yes" : "no")<<endl;}
#define PRINT_ARRAY_H(a) REPALL(__i,a) {cout<<a[__i]<<(__i != a.size()-1 ? " " : "");} newline;
#define PRINT_ARRAY_V(a) REPALL(__i,a) {cout<<a[__i]<<endl;}
#define e1 first
#define e2 second
#define newline putchar('\n')
#define cspace ' '
#define sspace " "
const int INF = 1e7;
const ll MOD = 1e9 + 7;
const double pi = 3.141592653589793;
const VI dx = {1, 0, -1, 0};
const VI dy = {0, 1, 0, -1};
//const VI dx = {1, 1, 0, -1, -1, -1, 0, 1};
//const VI dy = {0, 1, 1, 1, 0, -1, -1, -1};

//蟻本参考
//整数配列の和に対応
class BIT {
public:
    VLL bit;
    int num;

    BIT(int n) {
        num = n;
        bit.resize(n + 1);
    }

    ll sum(int i)
    {
        ll r = 0ll;
        while (i > 0) {
            r += bit[i];
            i -= i & -i;
        }
        return r;
    }

    void add(int i, ll x)
    {
        while (i <= num) {
            bit[i] += x;
            i += i & -i;
        }
    }

    void sub(int i, ll x)
    {
        add(i,-x);
    }
};

void solve()
{
    ll n, q;
    VLL c, ans, la; //last appear
    string s, t;
    cin>>n>>q;
    la.resize(n+1, -1);
    REP(i,n) {
        ll tc = input<ll>();
        c.push_back(tc);
    }
    ans.resize(q);
    vector<tuple<int, int, int>> queries;
    REP(i,q) {
        int l, r;
        cin>>l>>r;
        //l--; r--;
        queries.emplace_back(r, l, i);
    }
    SORT(queries);
    BIT bit(n);
    int cp = 0; //current position
    REP(i,q) {
        int l = get<1>(queries[i]);
        int r = get<0>(queries[i]);
        FOR(j,cp,r) {
            if (la[c[j]] == -1) {
                bit.add(j+1,1);
            } else {
                bit.sub(la[c[j]], 1);
                bit.add(j+1,1);
            }
            la[c[j]] = j + 1;
        }
        cp = r;
        ll tans = bit.sum(r);
        if (l > 0) tans -= bit.sum(l-1);

        ans[get<2>(queries[i])] = tans;
    }

    PRINT_ARRAY_V(ans);
}

int main()
{
    cout<<fixed<<setprecision(20);
    int __t = 1;
    //cin>>__t;
    REP(i,__t) {
        solve();
    }
    return 0;
}