codeforces 86D. Powerful array(莫队算法）

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 21 2 11 21 3

output

36

input

8 31 1 2 2 1 3 1 12 71 62 7

output

202020

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.

solution：

莫队算法模板题，只需要自己修改O（1）的部分

#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const int maxn = 220000;int unit;int num[1111111],a[maxn];long long tmp, ans[maxn];struct Query{int i, l, r;bool operator<(const Query &q)const{if (l / unit == q.l / unit) return r < q.r;return l / unit < q.l / unit;}}query[maxn];void insert(long long x){tmp -= x*num[x] * num[x];++num[x];tmp += x*num[x] * num[x];}void remove(long long x){tmp -= x*num[x] * num[x];--num[x];tmp += x*num[x] * num[x];}int main(){int n, t;scanf("%d%d", &n, &t);for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);unit = (int)sqrt((double)n);for (int i = 0; i < t; ++i){query[i].i = i;scanf("%d%d", &query[i].l, &query[i].r);}sort(query, query + t);int l = 1, r = 1; num[a[1]] = 1; tmp = a[1];for (int i = 0; i<t; ++i){while (l < query[i].l) remove(a[l++]);while (l > query[i].l) insert(a[--l]);while (r>query[i].r) remove(a[r--]);while (r < query[i].r) insert(a[++r]);ans[query[i].i] = tmp;}for (int i = 0; i < t; ++i)printf("%I64d\n", ans[i]);return 0;}