Problem: At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

  1. Print operation l, r. Picks should write down the value of <image>. 
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 
  3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x). 



Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.

Each of the next m lines begins with a number type <image>. 

  * If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1. 
  * If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2. 
  * If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3. 

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3


Output

8
5


Input

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


Output

49
15
23
1
9

Note

Consider the first testcase:

  * At first, a = {1, 2, 3, 4, 5}. 
  * After operation 1, a = {1, 2, 3, 0, 1}. 
  * After operation 2, a = {1, 2, 5, 0, 1}. 
  * At operation 3, 2 + 5 + 0 + 1 = 8. 
  * After operation 4, a = {1, 2, 2, 0, 1}. 
  * At operation 5, 1 + 2 + 2 = 5. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100 * 1000 + 10;
pair<int, int> seg_max[4 * MAX_N];
long long seg_sum[4 * MAX_N];
int n, m;
void modify(int p, int x, int id = 1, int l = 0, int r = n) {
  if (l == r - 1) {
    seg_sum[id] = x;
    seg_max[id] = {x, l};
    return;
  }
  int mid = (l + r) / 2;
  if (p < mid)
    modify(p, x, id * 2, l, mid);
  else
    modify(p, x, id * 2 + 1, mid, r);
  seg_sum[id] = seg_sum[id * 2] + seg_sum[id * 2 + 1];
  seg_max[id] = max(seg_max[id * 2], seg_max[id * 2 + 1]);
}
pair<int, int> get_max(int x, int y, int id = 1, int l = 0, int r = n) {
  if (x >= r || y <= l) return {0, 0};
  if (x <= l && y >= r) return seg_max[id];
  int mid = (l + r) / 2;
  return max(get_max(x, y, id * 2, l, mid), get_max(x, y, id * 2 + 1, mid, r));
}
long long get_sum(int x, int y, int id = 1, int l = 0, int r = n) {
  if (x >= r || y <= l) return 0;
  if (x <= l && y >= r) return seg_sum[id];
  int mid = (l + r) / 2;
  return get_sum(x, y, id * 2, l, mid) + get_sum(x, y, id * 2 + 1, mid, r);
}
void read_input() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    modify(i, a);
  }
}
void print_output() {
  for (int i = 0; i < m; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      l--;
      cout << get_sum(l, r) << endl;
    } else if (type == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      l--;
      pair<int, int> p = get_max(l, r);
      while (p.first >= x) {
        modify(p.second, p.first % x);
        p = get_max(l, r);
      }
    } else {
      int k, x;
      cin >> k >> x;
      k--;
      modify(k, x);
    }
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  read_input();
  print_output();
}