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http://www.elijahqi.win/archives/389D. 强大的数组时间限制：5秒内存限制：256MB输入：标准输入输出：标准输出给定一个正整数数组a1, a2, ..., an，请处理以下问题：

​​www.elijahqi.win/archives/389​​​
D. Powerful array
time limit per test
5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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
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.

这题最后输出要用i64d才行，lld会T掉。
题意求中各个数出现次数的平方×这个数的数值，其实就是小b的询问多加了点东西，这题卡常。。

#include<cstdio>
#include<cmath>
#include<algorithm>
#define LL long long
#define N 220000
inline int read(){
int x=0;char ch=getchar();
while (ch<'0'||ch>'9') ch=getchar();
while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
return x;
}
struct node{
int l,r,id,bl;
}q[N];
int n,t,n1;long long f[N<<2+1],a[N];
inline bool cmp(node a,node b){
return a.bl==b.bl?a.r<b.r:a.bl<b.bl;
}
long long ans[N];
int main(){
freopen("cf.in","r",stdin);
//freopen("cf.out","w",stdout);
n=read();t=read();n1=sqrt(n);
for (int i=1;i<=n;++i) a[i]=read();
for (int i=1;i<=t;++i) q[i].l=read(),q[i].r=read(),q[i].id=i,q[i].bl=(q[i].l/n1+1);
std::sort(q+1,q+t+1,cmp);
int l1=1,r1=0;long long tmp=0;
for (int i=1;i<=t;++i){
int l=q[i].l,r=q[i].r,id=q[i].id;
while (l1<l) f[a[l1]]--,tmp-=((f[a[l1]]<<1)+1)*a[l1],l1++;
while (l1>l) --l1,tmp+=((f[a[l1]]<<1)+1)*a[l1],f[a[l1]]++;
while (r<r1) f[a[r1]]--,tmp-=((f[a[r1]]<<1)+1)*a[r1],r1--;
while (r>r1) ++r1,tmp+=((f[a[r1]]<<1)+1)*a[r1],f[a[r1]]++;
ans[id]=tmp;
}
for (int i=1;i<=t;++i) printf("%I64d\n",ans[i]);
return 0;
}