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>
 
#define FOR(i, a, b) for(auto i=a; i<=b; ++i)
#define REP(i, a, b) for(auto i=a; i<b; ++i)
#define FORI(i, a, b) for(auto i=a; i!=b+1-2*(a>b); i+=1-2*(a>b))
#define REPI(i, a, b) for(auto i=a-(a>b); i!=b-(a>b); i+=1-2*(a>b))
#define ALL(v) v.begin(),v.end()
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)
#define pf(a) push_front(a)
#define eb(a, b) emplace_back(a, b)
#define fir first
#define sec second
#define what_is(x) cout<<#x<<" is "<<x<<endl;
#define type(x) typeid(x).name()
#define ms(arr, val) memset(arr, val, sizeof(arr))
#define min3(a,b,c) min(min(a,b),c)
#define max3(a,b,c) max(max(a,b),c)
#define PI acos(-1)
#define open_read freopen("christmas.in", "r", stdin)
#define open_write freopen("christmas.out", "w", stdout)
 
using namespace std;
 
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<double, double> PDD;
typedef pair<int, int> PII;
typedef pair<int, PII> PIPII;
typedef pair<PII, PII> PPIIPII;
typedef pair<LL, LL> PLL;
typedef pair<ULL, ULL> PUU;
typedef pair<LL, PLL> PLPLL;
typedef pair<PLL, PLL> PPLLPLL;
typedef pair<int, LL> PIL;
typedef pair<LL, int> PLI;
 
template<typename T, typename T1>
ostream& operator << (ostream &out, pair<T, T1> obj) {
    out << "(" << obj.first << ", " << obj.second << ")";
    return out;
}
 
template<typename T, typename T1>
ostream& operator << (ostream &out, map<T, T1> cont) {
    typename map<T, T1>::const_iterator itr = cont.begin();
    typename map<T, T1>::const_iterator ends = cont.end();
 
    for(; itr != ends; ++itr) {
        out << *itr << " ";
    }
 
    out<<endl;
    return out;
} 

template<typename T>
ostream& operator << (ostream &out, set<T> cont) {
    typename set<T>::const_iterator itr = cont.begin();
    typename set<T>::const_iterator ends = cont.end();
    
    for(; itr != ends; ++itr) {
        out << *itr << " ";
    }
 
    out << endl;
    return out;
}
  
template<typename T>
ostream& operator << (ostream &out, multiset<T> cont) {
    typename multiset<T>::const_iterator itr = cont.begin();
    typename multiset<T>::const_iterator ends = cont.end();
 
    for (; itr != ends; ++itr) {
        out << *itr << " ";
    }
 
    out << endl;
    return out;
}

template<typename T, template<typename ELEM, typename ALLOC=allocator<ELEM>> class CONT>
ostream& operator << (ostream &out, CONT<T> cont) {
    typename CONT<T>::const_iterator itr = cont.begin();
    typename CONT<T>::const_iterator ends = cont.end();
 
    for (; itr != ends; ++itr) {
        out << *itr << " ";
    }
 
    out << endl;
    return out;
}
 
template<typename T, unsigned int N, typename CTy, typename CTr>
typename enable_if<!is_same<T, char>::value, basic_ostream<CTy, CTr> &>::type
operator << (basic_ostream<CTy, CTr> &out, const T (&arr)[N]) {
    REP (i, 0, N) {
        out << arr[i] << " ";
    }
 
    out << endl;
    return out;
}
  
template<typename T>
T GCD(T a, T b) {
    T min_v = min(a, b);
    T max_v = max(a, b);
 
    while (min_v) {
        T temp = max_v % min_v;
        max_v = min_v;
        min_v = temp;
    }
 
    return max_v;
}
 
template<typename T>
T LCM(T a, T b) {
    return (a * b) / gcd(a, b);
}
 
template<typename T>
T fastExpPow(T base, T exp, T mod) {
    T res = 1;
 
    while (exp) {
        if (exp & 1) {
            res *= base;
            res %= mod;
        }
 
        exp >>= 1;
        base *= base;
        base %= mod;
    }
 
    return res % mod;
}
 
/*#################################################################################################################
###################################################################################################################
###################################################################################################################
#################################################################################################################*/

#define SIZE 500010

int N, Q;
int color[SIZE], BITree[SIZE], highPos[SIZE], result[SIZE];
vector<pair<PII, int>> queries;


int getSum(int index) { 
    if (index < 0) {
        return 0;
    }

    int sum = 0; 
    index = index + 1; 
  
    while (index > 0) { 
        sum += BITree[index]; 
        index -= index & (-index); 
    }

    return sum; 
} 

void updateBIT(int index, int val) { 
    index = index + 1; 
  
    while (index <= N) { 
        BITree[index] += val; 
        index += index & (-index); 
    } 
} 


int main() {
    scanf("%d%d", &N, &Q);
    
    REP (i, 0, N) {
        scanf("%d", color + i);
    }
    
    REP (i, 0, Q) {
        int L, R;
        scanf("%d%d", &L, &R);
        --L, --R;

        queries.eb(mp(R, L), i);
    }
    
    sort(ALL(queries));
    ms(highPos, -1);

    auto itr = queries.begin();

    REP (i, 0, N) {
        int col = color[i];

        if (highPos[col] != -1) {
            updateBIT(highPos[col], -1);
        }
        
        highPos[col] = i;
        updateBIT(highPos[col], 1);

        while (itr != queries.end() && itr->fir.fir == i) {
            PII seg = itr->fir;
            int sum = getSum(seg.fir) - getSum(seg.sec - 1);    

            result[itr->sec] = sum;
            ++itr;
        }
    }
    
    REP (i, 0, Q) {
        printf("%d\n", result[i]);
    }

    return 0;
}