9 10
6 2 3 1 4 8 5 9 8
3 9
6 6
9 9
9 9
9 9
2 3
3 9
3 8
6 9
5 6
 10 10
5 10 9 2 9 3 5 5 6 5
5 10
2 2
3 10
4 8
1 3
1 5
5 5
10 10
8 9
6 10
 8 10
100 100 100 100 100 100 100 100
1 8
2 8
3 8
4 8
5 8
6 8
7 8
8 8
1 1
1 5
 5 15
103 45 103 67 45
1 1
1 2
1 3
1 4
1 5
2 2
2 3
2 4
2 5
3 3
3 4
3 5
4 4
4 5
5 5
 1 1
1000000
1 1
 10 5
10 9 8 7 6 5 4 3 2 1
4 8
1 10
3 9
2 2
5 10
 9 10
5 3 10 2 4 8 7 5 1
6 9
1 8
3 3
2 7
3 5
4 6
9 9
8 8
1 9
5 6
 9 10
10 2 6 2 4 6 6 4 1
4 8
8 9
2 3
8 9
9 9
4 5
1 7
6 6
6 8
1 1
 20 8
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
4 15
1 2
2 20
7 7
13 18
7 7
3 19
3 8
 1 2
1
1 1
1 1
 8 10
5 7 3 1 4 2 1 1
5 7
3 8
2 7
4 8
4 5
2 8
4 6
1 2
4 7
3 5

54
8
8
8
8
5
54
30
46
12
 63
10
92
34
24
53
9
5
11
54
 6400
4900
3600
2500
1600
900
400
100
100
2500
 103
148
457
524
659
45
148
215
350
103
170
215
67
112
45
 1000000
 25
55
35
9
21
 21
54
10
34
16
14
1
5
55
12
 42
5
8
5
1
6
76
6
28
10
 108
3
281
1
27
1
209
27
 1
1
 7
18
20
15
5
25
7
12
10
8

#include <bits/stdc++.h>
using namespace std;
using lli = long long int;
using pii = pair<int, int>;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vlli = vector<long long int>;
using vpii = vector<pair<int, int>>;
const int N = 200005;
const int blockSize = sqrt(N);
lli n, t, l, r, x, idx, L, R, res;
vlli a, ans, cnt(1000005);
vector<tuple<lli, lli, lli, lli>> q;
int get(int idx) { return idx / blockSize; }
void add(lli x) {
  res += x * (2 * cnt[x] + 1);
  cnt[x]++;
}
void del(lli x) {
  res += x * (-2 * cnt[x] + 1);
  cnt[x]--;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> t;
  a.resize(n);
  ans.resize(t);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < t; i++) {
    cin >> l >> r;
    l--;
    r--;
    q.push_back(make_tuple(get(l), r, l, i));
  }
  sort((q).begin(), (q).end());
  L = R = 0;
  add(a[0]);
  for (int i = 0; i < t; i++) {
    tie(ignore, r, l, idx) = q[i];
    while (L < l) del(a[L++]);
    while (L > l) add(a[--L]);
    while (R < r) add(a[++R]);
    while (R > r) del(a[R--]);
    ans[idx] = res;
  }
  for (int i = 0; i < t; i++) cout << ans[i] << '\n';
}
