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;
template <class T>
inline void read(T& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
inline char getc() {
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  return c;
}
const int N = 200010;
int n, m, i, j, p, ki;
bool fir;
long long o[N], x[N], X[N], a[N][3];
struct node {
  long long sx, g, res;
} T[N * 4], ans;
node merge(node a, node b) {
  node c;
  c.g = a.g + b.g;
  c.sx = a.sx + b.sx;
  c.res = a.res + b.res;
  c.res += b.sx * a.g - a.sx * b.g;
  return c;
}
void find(int i, int x, int y, int l, int r) {
  if (l > r) return;
  if (y < l || x > r) return;
  if (x >= l && y <= r) {
    if (ki == 0)
      T[i] = (node){0, 0, 0};
    else if (ki == 1)
      T[i] = (node){o[x], 1, 0};
    else if (ki == 2) {
      if (fir)
        fir = 0, ans = T[i];
      else
        ans = merge(ans, T[i]);
    }
    return;
  }
  find(i * 2, x, (x + y) / 2, l, r);
  find(i * 2 + 1, (x + y) / 2 + 1, y, l, r);
  T[i] = merge(T[i * 2], T[i * 2 + 1]);
}
int F(long long x) { return lower_bound(o + 1, o + 1 + o[0], x) - o; }
int get(int i, int x, int y, int K) {
  if (x == y) return x;
  if (K <= T[i * 2].g)
    return get(i * 2, x, (x + y) / 2, K);
  else
    return get(i * 2 + 1, (x + y) / 2 + 1, y, K - T[i * 2].g);
}
int main() {
  read(n);
  for (i = 1; i <= n; i++) read(x[i]), o[++o[0]] = x[i];
  memcpy(X, x, sizeof(x));
  read(m);
  for (i = 1; i <= m; i++) read(a[i][0]), read(a[i][1]), read(a[i][2]);
  for (i = 1; i <= m; i++)
    if (a[i][0] == 1) {
      x[a[i][1]] += a[i][2];
      o[++o[0]] = x[a[i][1]];
    }
  sort(o + 1, o + 1 + o[0]);
  j = 1;
  for (i = 2; i <= o[0]; i++)
    if (o[i] != o[i - 1]) o[++j] = o[i];
  o[0] = j;
  memcpy(x, X, sizeof(X));
  for (i = 1; i <= n; i++) {
    ki = 1;
    find(1, 1, o[0], F(x[i]), F(x[i]));
  }
  for (i = 1; i <= m; i++) {
    if (a[i][0] == 1) {
      int u = a[i][1], v = a[i][2];
      ki = 0;
      find(1, 1, o[0], F(x[u]), F(x[u]));
      x[u] += v;
      ki = 1;
      find(1, 1, o[0], F(x[u]), F(x[u]));
    } else {
      int x, y;
      x = a[i][1];
      y = a[i][2];
      x = lower_bound(o + 1, o + 1 + o[0], x) - o;
      y = upper_bound(o + 1, o + 1 + o[0], y) - o - 1;
      fir = 1;
      ki = 2;
      find(1, 1, o[0], x, y);
      if (fir)
        printf("0\n");
      else
        printf("%I64d\n", ans.res);
    }
  }
  scanf("\n");
  return 0;
}
