#include <bits/stdc++.h>
using namespace std;
int n, a[1000006], d[1000006], c[1000006];
void build(int v, int l, int r) {
  if (l == r) {
    d[v] = a[l];
    return;
  }
  int m = (l + r) / 2;
  build(v * 2, l, m);
  build(v * 2 + 1, m + 1, r);
}
void push(int v) {
  c[v * 2 + 1] += c[v];
  c[v * 2] += c[v];
  c[v] = 0;
}
void update(int v, int l, int r, int pos, int val) {
  int m = (l + r) / 2;
  if (l == r) {
    d[v] = val;
    c[v] = 0;
    return;
  }
  if (c[v] > 0) {
    push(v);
  }
  if (pos <= m)
    update(v * 2, l, m, pos, val);
  else
    update(v * 2 + 1, m + 1, r, pos, val);
}
int get(int v, int l, int r, int pos) {
  if (l == r) {
    return c[v] + d[v];
  }
  if (c[v] > 0) {
    push(v);
  }
  int m = (l + r) / 2;
  if (pos <= m)
    return get(v * 2, l, m, pos);
  else
    return get(v * 2 + 1, m + 1, r, pos);
}
int main() {
  int n, m, t, v, x, n2 = 1, q[100005], size = 0;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  while (n2 < n) n2 *= 2;
  build(1, 1, n2);
  for (int i = 0; i < m; i++) {
    cin >> t;
    if (t == 1) {
      cin >> v >> x;
      update(1, 1, n2, v, x);
    }
    if (t == 2) {
      cin >> x;
      c[1] += x;
    }
    if (t == 3) {
      cin >> v;
      q[size++] = get(1, 1, n2, v);
    }
  }
  for (int i = 0; i < size; i++) cout << q[i] << endl;
}

### Prompt

Generate a CPP solution to the following problem:

Sereja has got an array, consisting of n integers, a1, a2, ..., an. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:

  1. Make vi-th array element equal to xi. In other words, perform the assignment avi = xi. 
  2. Increase each array element by yi. In other words, perform n assignments ai = ai + yi (1 ≤ i ≤ n). 
  3. Take a piece of paper and write out the qi-th array element. That is, the element aqi. 



Help Sereja, complete all his operations.

Input

The first line contains integers n, m (1 ≤ n, m ≤ 105). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the original array.

Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer ti (1 ≤ ti ≤ 3) that represents the operation type. If ti = 1, then it is followed by two integers vi and xi, (1 ≤ vi ≤ n, 1 ≤ xi ≤ 109). If ti = 2, then it is followed by integer yi (1 ≤ yi ≤ 104). And if ti = 3, then it is followed by integer qi (1 ≤ qi ≤ n).

Output

For each third type operation print value aqi. Print the values in the order, in which the corresponding queries follow in the input.

Examples

Input

10 11
1 2 3 4 5 6 7 8 9 10
3 2
3 9
2 10
3 1
3 10
1 1 10
2 10
2 10
3 1
3 10
3 9


Output

2
9
11
20
30
40
39

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, a[1000006], d[1000006], c[1000006];
void build(int v, int l, int r) {
  if (l == r) {
    d[v] = a[l];
    return;
  }
  int m = (l + r) / 2;
  build(v * 2, l, m);
  build(v * 2 + 1, m + 1, r);
}
void push(int v) {
  c[v * 2 + 1] += c[v];
  c[v * 2] += c[v];
  c[v] = 0;
}
void update(int v, int l, int r, int pos, int val) {
  int m = (l + r) / 2;
  if (l == r) {
    d[v] = val;
    c[v] = 0;
    return;
  }
  if (c[v] > 0) {
    push(v);
  }
  if (pos <= m)
    update(v * 2, l, m, pos, val);
  else
    update(v * 2 + 1, m + 1, r, pos, val);
}
int get(int v, int l, int r, int pos) {
  if (l == r) {
    return c[v] + d[v];
  }
  if (c[v] > 0) {
    push(v);
  }
  int m = (l + r) / 2;
  if (pos <= m)
    return get(v * 2, l, m, pos);
  else
    return get(v * 2 + 1, m + 1, r, pos);
}
int main() {
  int n, m, t, v, x, n2 = 1, q[100005], size = 0;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  while (n2 < n) n2 *= 2;
  build(1, 1, n2);
  for (int i = 0; i < m; i++) {
    cin >> t;
    if (t == 1) {
      cin >> v >> x;
      update(1, 1, n2, v, x);
    }
    if (t == 2) {
      cin >> x;
      c[1] += x;
    }
    if (t == 3) {
      cin >> v;
      q[size++] = get(1, 1, n2, v);
    }
  }
  for (int i = 0; i < size; i++) cout << q[i] << endl;
}

```