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 long double pi = acos(-1);
const int n_ = 1e5 + 5;
int n, q, t, arr[n_];
pair<long long, long long> st[n_ << 2];
void build(int id = 1, int l = 1, int r = n) {
  if (l == r) {
    st[id] = {arr[l], arr[l]};
    return;
  }
  build(id << 1, l, ((l + r) >> 1));
  build(id << 1 | 1, ((l + r) >> 1) + 1, r);
  st[id] = {st[id << 1].first + st[id << 1 | 1].first,
            max(st[id << 1].second, st[id << 1 | 1].second)};
}
void modupd(int s, int en, int val, int id = 1, int l = 1, int r = n) {
  if (st[id].second < val || l > en || r < s) return;
  if (l == r) {
    int x = st[id].first % val;
    st[id] = {x, x};
    return;
  }
  modupd(s, en, val, id << 1, l, ((l + r) >> 1));
  modupd(s, en, val, id << 1 | 1, ((l + r) >> 1) + 1, r);
  st[id] = {st[id << 1].first + st[id << 1 | 1].first,
            max(st[id << 1].second, st[id << 1 | 1].second)};
}
void upd(int x, int val, int id = 1, int l = 1, int r = n) {
  if (x > r || x < l) return;
  if (l == r) {
    st[id] = {val, val};
    return;
  }
  upd(x, val, id << 1, l, ((l + r) >> 1));
  upd(x, val, id << 1 | 1, ((l + r) >> 1) + 1, r);
  st[id] = {st[id << 1].first + st[id << 1 | 1].first,
            max(st[id << 1].second, st[id << 1 | 1].second)};
}
long long query(int s, int e, int id = 1, int l = 1, int r = n) {
  if (s > r || e < l) return 0;
  if (s <= l && r <= e) return st[id].first;
  return query(s, e, id << 1, l, ((l + r) >> 1)) +
         query(s, e, id << 1 | 1, ((l + r) >> 1) + 1, r);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  build();
  int l, r, x;
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d %d", &l, &r);
      printf("%lld\n", query(l, r));
    } else if (t == 2) {
      scanf("%d %d %d", &l, &r, &x);
      modupd(l, r, x);
    } else {
      scanf("%d %d", &l, &r);
      upd(l, r);
    }
  }
  return 0;
}