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 N = 1e5 + 1;
struct Node {
  long long mx, sum, idmx;
  void mergee(Node a, Node b) {
    mx = max(a.mx, b.mx);
    sum = a.sum + b.sum;
    if (a.mx >= b.mx)
      idmx = a.idmx;
    else
      idmx = b.idmx;
  }
} ST[4 * N];
long long n, m, a[N];
void update(int id, int l, int r, int i, int v) {
  if (i < l || i > r) return;
  if (r == l) {
    ST[id].mx = ST[id].sum = v;
    ST[id].idmx = i;
    a[i] = v;
    return;
  }
  int mid = l + r >> 1;
  update(id * 2, l, mid, i, v);
  update(id * 2 + 1, mid + 1, r, i, v);
  ST[id].mergee(ST[id * 2], ST[id * 2 + 1]);
}
Node get(int id, int l, int r, int u, int v) {
  Node res = {0, 0, 0};
  if (r < u || l > v) return res;
  if (u <= l && r <= v) return ST[id];
  int mid = l + r >> 1;
  res.mergee(get(id * 2, l, mid, u, v), get(id * 2 + 1, mid + 1, r, u, v));
  return res;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) update(1, 1, n, i, a[i]);
  while (m--) {
    int type, l, r, x, k;
    cin >> type;
    if (type == 1) {
      cin >> l >> r;
      cout << get(1, 1, n, l, r).sum << endl;
    } else if (type == 2) {
      cin >> l >> r >> x;
      while (get(1, 1, n, l, r).mx >= x)
        update(1, 1, n, get(1, 1, n, l, r).idmx,
               a[get(1, 1, n, l, r).idmx] % x);
    } else
      cin >> k >> x, update(1, 1, n, k, x);
  }
}