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 len = 1 << 17;
int n, m, x, y, k;
long long second[300300];
pair<int, int> a[300300];
void upd(int pos, int val) {
  a[pos].first = val;
  second[pos] = val;
  while (pos > 1) {
    pos /= 2;
    if (a[pos * 2].first > a[pos * 2 + 1].first)
      a[pos] = a[pos * 2];
    else
      a[pos] = a[pos * 2 + 1];
    second[pos] = second[pos * 2] + second[pos * 2 + 1];
  }
}
pair<int, int> find_max(int l, int r) {
  pair<int, int> ans;
  ans.first = 0;
  while (l <= r) {
    if (l % 2 == 1 && ans.first < a[l].first) ans = a[l];
    if (r % 2 == 0 && ans.first < a[r].first) ans = a[r];
    l = (l + 1) / 2;
    r = (r - 1) / 2;
  }
  return ans;
}
long long sum(int l, int r) {
  long long res = 0;
  while (l <= r) {
    if (l % 2 == 1) res += second[l];
    if (r % 2 == 0) res += second[r];
    l = (l + 1) / 2;
    r = (r - 1) / 2;
  }
  return res;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> x;
    a[len + i].second = i;
    upd(len + i, x);
  }
  for (int i = 0; i < m; ++i) {
    cin >> k;
    if (k == 1) {
      cin >> x >> y;
      cout << sum(x + len - 1, y + len - 1) << endl;
    } else if (k == 2) {
      cin >> x >> y >> k;
      while (true) {
        pair<int, int> mx = find_max(x + len - 1, y + len - 1);
        if (mx.first < k) break;
        upd(mx.second + len, mx.first % k);
      }
    } else {
      cin >> x >> y;
      upd(x + len - 1, y);
    }
  }
  return 0;
}