Yaroslav has n points that lie on the Ox axis. The coordinate of the first point is x1, the coordinate of the second point is x2, ..., the coordinate of the n-th point is — xn. Now Yaroslav wants to execute m queries, each of them is of one of the two following types:

  1. Move the pj-th point from position xpj to position xpj + dj. At that, it is guaranteed that after executing such query all coordinates of the points will be distinct. 
  2. Count the sum of distances between all pairs of points that lie on the segment [lj, rj] (lj ≤ rj). In other words, you should count the sum of: <image>. 



Help Yaroslav.

Input

The first line contains integer n — the number of points (1 ≤ n ≤ 105). The second line contains distinct integers x1, x2, ..., xn — the coordinates of points (|xi| ≤ 109).

The third line contains integer m — the number of queries (1 ≤ m ≤ 105). The next m lines contain the queries. The j-th line first contains integer tj (1 ≤ tj ≤ 2) — the query type. If tj = 1, then it is followed by two integers pj and dj (1 ≤ pj ≤ n, |dj| ≤ 1000). If tj = 2, then it is followed by two integers lj and rj ( - 109 ≤ lj ≤ rj ≤ 109).

It is guaranteed that at any moment all the points have distinct coordinates.

Output

For each type 2 query print the answer on a single line. Print the answers in the order, in which the queries follow in the input.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Examples

Input

8
36 50 28 -75 40 -60 -95 -48
20
2 -61 29
1 5 -53
1 1 429
1 5 130
2 -101 -71
2 -69 53
1 1 404
1 5 518
2 -101 53
2 50 872
1 1 -207
2 -99 -40
1 7 -389
1 6 -171
1 2 464
1 7 -707
1 1 -730
1 1 560
2 635 644
1 7 -677


Output

176
20
406
1046
1638
156
0
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;
int n, m, n1, cord[N], p[N << 1], cur[N], q[N][3];
struct info {
  int c;
  long long sum, ans;
  info(int _c = 0, long long _sum = 0, long long _ans = 0) {
    c = _c;
    sum = _sum;
    ans = _ans;
  }
} b[N << 3];
info merge(info a, info b) {
  return info(a.c + b.c, a.sum + b.sum,
              a.ans + b.ans + a.c * b.sum - b.c * a.sum);
}
void update(int node, int L, int R, int x, int v) {
  if (p[L] > cord[x] || p[R] < cord[x]) return;
  if (L == R) {
    if (v == 1)
      b[node] = info(1, cord[x], 0);
    else
      b[node] = info();
    return;
  }
  update(node * 2, L, (L + R) / 2, x, v);
  update(node * 2 + 1, (L + R) / 2 + 1, R, x, v);
  b[node] = merge(b[node * 2], b[node * 2 + 1]);
}
info get(int node, int L, int R, int l, int r) {
  if (L > r || R < l) return info();
  if (l <= L && R <= r) return b[node];
  return merge(get(node * 2, L, (L + R) / 2, l, r),
               get(node * 2 + 1, (L + R) / 2 + 1, R, l, r));
}
void prepare() {
  scanf("%d", &n);
  for (auto i = 1; i <= n; i++) scanf("%d", cord + i);
  n1 = n;
  for (auto i = 1; i <= n; i++) p[i] = cur[i] = cord[i];
  scanf("%d", &m);
  for (auto i = 1; i <= m; i++) {
    for (auto j = 0; j <= 2; j++) scanf("%d", q[i] + j);
    if (q[i][0] == 1) {
      cur[q[i][1]] += q[i][2];
      p[++n1] = cur[q[i][1]];
    }
  }
  sort(p + 1, p + n1 + 1);
  n1 = unique(p + 1, p + n1 + 1) - p - 1;
  for (auto i = 1; i <= n; i++) update(1, 1, n1, i, 1);
}
void check(int node, int L, int R) {
  printf("%d %d %d %d %lld\n", node, L, R, b[node].c, b[node].sum);
  if (L < R) {
    check(node * 2, L, (L + R) / 2);
    check(node * 2 + 1, (L + R) / 2 + 1, R);
  }
}
void solve() {
  for (auto i = 1; i <= m; i++) {
    if (q[i][0] == 1) {
      update(1, 1, n1, q[i][1], -1);
      cord[q[i][1]] += q[i][2];
      update(1, 1, n1, q[i][1], 1);
    } else {
      int L = lower_bound(p + 1, p + n1 + 1, q[i][1]) - p,
          R = upper_bound(p + 1, p + n1 + 1, q[i][2]) - p - 1;
      printf("%I64d\n", get(1, 1, n1, L, R).ans);
    }
  }
}
int main() {
  prepare();
  solve();
}
