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;
const int N = 2e5 + 10;
const int M = 1e6 + 10;
int a[N], cnt[M];
long long ans[N], res;
struct Q {
  int l, r, b, id;
  Q(){};
  bool operator<(const Q& rhs) const {
    if (b == rhs.b) return r < rhs.r;
    return b < rhs.b;
  }
} q[N];
int L, R, first;
long long cal1(int l, int r) {
  long long res = 0;
  for (int i = l; i <= r; i++) {
    res += 1LL * a[i] * (2 * cnt[a[i]] + 1);
    cnt[a[i]]++;
  }
  return res;
}
long long cal2(int l, int r) {
  long long res = 0;
  for (int i = l; i <= r; i++) {
    cnt[a[i]]--;
    res -= 1LL * a[i] * (2 * cnt[a[i]] + 1);
  }
  return res;
}
long long query(int x, int y) {
  if (first) {
    first = 0;
    res = cal1(x, y);
  } else {
    res += cal1(x, L - 1);
    res += cal2(L, x - 1);
    res += cal1(R + 1, y);
    res += cal2(y + 1, R);
  }
  L = x, R = y;
  return res;
}
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int block_size = sqrt(n + 0.5);
  for (int i = 0; i < t; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].b = q[i].l / block_size, q[i].id = i;
  }
  sort(q, q + t);
  memset(cnt, 0, sizeof(cnt));
  first = 1;
  res = 0;
  for (int i = 0; i < t; i++) ans[q[i].id] = query(q[i].l, q[i].r);
  for (int i = 0; i < t; i++) printf("%lld\n", ans[i]);
  return 0;
}
