Codeforces#86D Powerful array（莫队）_an array of positive integers let us consider-优快云博客

本文链接：https://blog.youkuaiyun.com/dsaghjkye/article/details/82289496

本文介绍了一个算法问题，即计算给定数组中多个子数组的能量值总和。能量定义为子数组中每个唯一元素频率的平方乘以其值的总和。文章提供了一种高效的算法实现方案，并通过具体例子说明了计算过程。

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An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occurrences of s into the subarray. We call the power of the subarray the sum of products Ks·Ks·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 ai (1 ≤ ai ≤ 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

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.

#include<map>
#include<vector>
#include<iostream>
#include<queue>
#include<deque>
#include<math.h>
#include<algorithm>
#include<stdio.h>
#include<string>
#include<string.h>
#define ll long long
const int maxn=1e6+10;
using namespace std;
int a[maxn],sum[maxn];
int t;
ll anss,ans[maxn];
struct node
{
    int id,l,r;
}e[maxn];
int cmp(node n1,node n2)
{
    if(n1.l/t==n2.l/t)
        return n1.r<n2.r;
    return n1.l<n2.l;
}
void aa(int x,int add)
{
    anss-=(1ll*sum[x]*sum[x]*x);
    sum[x]+=add;
    anss+=(1ll*sum[x]*sum[x]*x);
}
int main()
{
    anss=0;
    int n,m,l,r;
    scanf("%d%d",&n,&m);
    memset(sum,0,sizeof(sum));
     t=(int)sqrt(1.0*n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&e[i].l,&e[i].r);
        e[i].l--;
        e[i].r--;
        e[i].id=i;
    }
    sort(e,e+m,cmp);
    l=0,r=-1;
    for(int i=0;i<m;i++)
    {
        while(e[i].l>l) aa(a[l++],-1);
        while(e[i].l<l) aa(a[--l],1);
        while(e[i].r>r) aa(a[++r],1);
        while(e[i].r<r) aa(a[r--],-1);
        ans[e[i].id]=anss;
    }
    for(int i=0;i<m;i++)
        printf("%I64d\n",ans[i]);

}