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;
char buf[1 << 21], *p1 = buf, *p2 = buf;
template <class T>
void read(T &x) {
  x = 0;
  int c = getchar();
  int flag = 0;
  while (c < '0' || c > '9') flag |= (c == '-'), c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (flag) x = -x;
}
template <class T>
T _max(T a, T b) {
  return b < a ? a : b;
}
template <class T>
T _min(T a, T b) {
  return a < b ? a : b;
}
template <class T>
bool checkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
bool checkmin(T &a, T b) {
  return b < a ? a = b, 1 : 0;
}
const int N = 100005, M = 500005;
int n, m;
int a[N], A[N];
int arr[M], top = 0;
int typ[N], c1[N], c2[N];
int cnt[M];
struct node {
  int cnt;
  long long sum, total;
} t[M << 2];
node operator+(node a, node b) {
  node c;
  c.cnt = a.cnt + b.cnt;
  c.sum = a.sum + b.sum;
  c.total = a.total + b.total + a.cnt * b.sum - b.cnt * a.sum;
  return c;
}
void push_up(int x) { t[x] = t[x << 1] + t[x << 1 | 1]; }
void build(int x, int l, int r) {
  if (l == r) {
    t[x].cnt = cnt[l];
    t[x].sum = (long long)cnt[l] * arr[l];
    t[x].total = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(x << 1, l, mid);
  build(x << 1 | 1, mid + 1, r);
  push_up(x);
}
void update(int x, int l, int r, int pos, int d) {
  if (l == r) {
    t[x].sum += (long long)d * arr[l];
    t[x].cnt += d;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    update(x << 1, l, mid, pos, d);
  else
    update(x << 1 | 1, mid + 1, r, pos, d);
  push_up(x);
}
node query(int x, int l, int r, int L, int R) {
  if (L <= l && r <= R) {
    return t[x];
  }
  int mid = (l + r) >> 1;
  if (R <= mid) return query(x << 1, l, mid, L, R);
  if (L > mid) return query(x << 1 | 1, mid + 1, r, L, R);
  return query(x << 1, l, mid, L, R) + query(x << 1 | 1, mid + 1, r, L, R);
}
void init() {
  read(n);
  for (int i = 1; i <= n; ++i) read(a[i]), arr[++top] = a[i], A[i] = a[i];
  read(m);
  for (int i = 1; i <= m; ++i) {
    read(typ[i]), read(c1[i]), read(c2[i]);
    if (typ[i] == 1) {
      A[c1[i]] += c2[i];
      arr[++top] = A[c1[i]];
    } else {
      arr[++top] = c1[i];
      arr[++top] = c2[i];
    }
  }
  sort(arr + 1, arr + top + 1);
  top = unique(arr + 1, arr + top + 1) - arr - 1;
  for (int i = 1; i <= n; ++i)
    ++cnt[lower_bound(arr + 1, arr + top + 1, a[i]) - arr];
  build(1, 1, top);
}
void solve() {
  int t;
  for (int i = 1; i <= m; ++i) {
    if (typ[i] == 1) {
      int idx = c1[i];
      t = lower_bound(arr + 1, arr + top + 1, a[idx]) - arr;
      update(1, 1, top, t, -1);
      a[idx] += c2[i];
      t = lower_bound(arr + 1, arr + top + 1, a[idx]) - arr;
      update(1, 1, top, t, 1);
    } else {
      int l = lower_bound(arr + 1, arr + top + 1, c1[i]) - arr;
      int r = lower_bound(arr + 1, arr + top + 1, c2[i]) - arr;
      node t = query(1, 1, top, l, r);
      printf("%lld\n", t.total);
    }
  }
}
int main() {
  init();
  solve();
  return 0;
}
