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;
struct nd {
  int mx, ind;
  long long sum;
};
int n, m, a[100005], t, p1, p2, p3;
nd seg[270000];
void build(int l, int r, int cur) {
  if (l == r) {
    seg[cur] = {a[l], l, a[l]};
    return;
  }
  build(l, (l + r) / 2, 2 * cur + 1);
  build((l + r) / 2 + 1, r, 2 * cur + 2);
  seg[cur].mx = max(seg[2 * cur + 1].mx, seg[2 * cur + 2].mx);
  seg[cur].sum = seg[2 * cur + 1].sum + seg[2 * cur + 2].sum;
  if (seg[2 * cur + 1].mx >= seg[2 * cur + 2].mx)
    seg[cur].ind = seg[2 * cur + 1].ind;
  else
    seg[cur].ind = seg[2 * cur + 2].ind;
}
void upd(int l, int r, int targ, int cur, int val) {
  if (r < targ or targ < l) return;
  if (l == r) {
    seg[cur] = {val, l, val};
    return;
  }
  upd(l, (l + r) / 2, targ, 2 * cur + 1, val);
  upd((l + r) / 2 + 1, r, targ, 2 * cur + 2, val);
  seg[cur].mx = max(seg[2 * cur + 1].mx, seg[2 * cur + 2].mx);
  seg[cur].sum = seg[2 * cur + 1].sum + seg[2 * cur + 2].sum;
  if (seg[2 * cur + 1].mx >= seg[2 * cur + 2].mx)
    seg[cur].ind = seg[2 * cur + 1].ind;
  else
    seg[cur].ind = seg[2 * cur + 2].ind;
}
long long que(int l, int r, int tl, int tr, int cur) {
  if (r < tl or tr < l) return 0;
  if (tl <= l and r <= tr) return seg[cur].sum;
  return que(l, (l + r) / 2, tl, tr, 2 * cur + 1) +
         que((l + r) / 2 + 1, r, tl, tr, 2 * cur + 2);
}
int getmax(int l, int r, int tl, int tr, int cur) {
  if (r < tl or tr < l) return 0;
  if (tl <= l and r <= tr) return seg[cur].ind;
  int i1 = getmax(l, (l + r) / 2, tl, tr, 2 * cur + 1),
      i2 = getmax((l + r) / 2 + 1, r, tl, tr, 2 * cur + 2);
  if (a[i1] > a[i2])
    return i1;
  else
    return i2;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(0, n, 0);
  for (int i = 0; i < m; i++) {
    cin >> t;
    if (t == 1) {
      cin >> p1 >> p2;
      cout << que(0, n, p1, p2, 0) << '\n';
    } else if (t == 2) {
      cin >> p1 >> p2 >> p3;
      int ind = getmax(0, n, p1, p2, 0);
      while (a[ind] >= p3) {
        upd(0, n, ind, 0, a[ind] % p3);
        a[ind] %= p3;
        ind = getmax(0, n, p1, p2, 0);
      }
    } else {
      cin >> p1 >> p2;
      upd(0, n, p1, 0, p2);
      a[p1] = p2;
    }
  }
}