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.

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 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).

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.

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e5+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;
///   数组大小
int pos[N],k,a[N],ji[M];
struct is
{int l,r,p;bool operator <(const is &b)const{if(pos[l]==pos[b.l])return r<b.r;return pos[l]<pos[b.l];}
}s[N];
ll ans;
void add(int x)
{ans-=1LL*ji[a[x]]*ji[a[x]]*a[x];ji[a[x]]++;ans+=1LL*ji[a[x]]*ji[a[x]]*a[x];
}
void del(int x)
{ans-=1LL*ji[a[x]]*ji[a[x]]*a[x];ji[a[x]]--;ans+=1LL*ji[a[x]]*ji[a[x]]*a[x];
}
ll out[N];
int main()
{int n,q;scanf("%d%d",&n,&q);k=sqrt(n);for(int i=1;i<=n;i++)scanf("%d",&a[i]),pos[i]=(i-1)/k+1;for(int i=1;i<=q;i++)scanf("%d%d",&s[i].l,&s[i].r),s[i].p=i;sort(s+1,s+1+q);int L=1,R=0;for(int i=1;i<=q;i++){while(L<s[i].l){del(L);L++;}while(L>s[i].l){L--;add(L);}while(R>s[i].r){del(R);R--;}while(R<s[i].r){R++;add(R);}out[s[i].p]=ans;}for(int i=1;i<=q;i++)printf("%lld\n",out[i]);return 0;
}