codeforces 86D : Powerful array

Description

An array of positive integers a₁, a₂, ..., a_n is given. Let us consider its arbitrary subarray a_l, a_l + 1..., a_r, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by K_s the number of occurrences of s into the subarray. We call the power of the subarray the sum of products K_s·K_s·s for every positive integer s. The sum contains only finite number of nonzero summands as the number of different values in the array is indeed finite.

You should calculate the power of t given subarrays.

Input

First line contains two integers n and t (1 ≤ n, t ≤ 200000) — the array length and the number of queries correspondingly.

Second line contains n positive integers a_i (1 ≤ a_i ≤ 10⁶) — the elements of the array.

Next t lines contain two positive integers l, r (1 ≤ l ≤ r ≤ n) each — the indices of the left and the right ends of the corresponding subarray.

Output

Output t lines, the i-th line of the output should contain single positive integer — the power of the i-th query subarray.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preferred to use cout stream (also you may use %I64d).

Examples

Input

3 2
1 2 1
1 2
1 3

Output

3
6

Input

8 3
1 1 2 2 1 3 1 1
2 7
1 6
2 7

Output

20
20
20

Note

Consider the following array (see the second sample) and its [2, 7] subarray (elements of the subarray are colored):

Then K₁ = 3, K₂ = 2, K₃ = 1, so the power is equal to 3²·1 + 2²·2 + 1²·3 = 20.

 

 

 

正解：莫队算法

解题报告：

　　莫队裸题，很好处理，感觉没什么好说的。

　　这别codeforces坑死了，只能用cout。。。严重拖慢我的速度。

 

//It is made by jump~
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
#ifdef WIN32   
#define OT "%I64d"
#else
#define OT "%lld"
#endif
using namespace std;
typedef long long LL;
const int MAXN = 200011;
const int size = 1000011;
int n,m,a[MAXN];
int cnt[size];
int kuai;
int nowl,nowr;
LL ans;
LL out[MAXN];

struct wen{
    int l,r,id,belong;
}q[MAXN];

inline int getint()
{
       int w=0,q=0;
       char c=getchar();
       while((c<'0' || c>'9') && c!='-') c=getchar();
       if (c=='-')  q=1, c=getchar();
       while (c>='0' && c<='9') w=w*10+c-'0', c=getchar();
       return q ? -w : w;
}

inline bool cmp(wen p,wen pp){ if(pp.belong==p.belong) return p.r<pp.r; return p.belong<pp.belong; }

inline bool ccmp(wen p,wen pp){ return p.id<pp.id; }

inline void add(int x,int type){
    if(type==1) {
    ans+=(LL)(cnt[a[x]]*2+1)*a[x];
    cnt[a[x]]++;
    }
    else{
    cnt[a[x]]--;
    ans-=(LL)(cnt[a[x]]*2+1)*a[x];
    }
}

inline void work(){
    n=getint(); m=getint(); kuai=sqrt(n); 
    for(int i=1;i<=n;i++) a[i]=getint();
    for(int i=1;i<=m;i++) q[i].l=getint(),q[i].r=getint(),q[i].id=i,q[i].belong=(q[i].l-1)/kuai+1;
    sort(q+1,q+m+1,cmp);
    nowl=q[1].l; nowr=q[1].r;
    for(int i=q[1].l;i<=q[1].r;i++) {
    //ans-=(LL)cnt[a[i]]*cnt[a[i]]*a[i];
    ans+=(LL)(cnt[a[i]]*2+1)*a[i];//简化
    cnt[a[i]]++;
    //ans+=(LL)cnt[a[i]]*cnt[a[i]]*a[i];
    }
    out[q[1].id]=ans;
    //注意nowl、nowr是当前位置，处理完毕的状态
    for(int i=2;i<=m;i++) {
    while(nowl<q[i].l) add(nowl,-1),nowl++;
    while(nowl>q[i].l) add(nowl-1,1),nowl--;//nowl已经插入，需要插入nowl-1
    while(nowr<q[i].r) add(nowr+1,1),nowr++;
    while(nowr>q[i].r) add(nowr,-1),nowr--;
    out[q[i].id]=ans;
    }
    for(int i=1;i<=m;i++) cout<<out[i]<<endl;
}

int main()
{
  work();
  return 0;
}