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 = 100010;
struct node {
  int id, m, v;
  long long s;
};
node nul;
node a[N << 2];
int n, m;
void update(int node, int l, int r, int il, int ir, int x) {
  if (il <= l && r <= ir) {
    a[node].id = l;
    a[node].m = x;
    a[node].v = x;
    a[node].s = 1LL * x * (r - l + 1);
    return;
  }
  if (a[node].v != 0) {
    a[node * 2].v = a[node].v;
    a[node * 2 + 1].v = a[node].v;
    a[node].v = 0;
  }
  int m = (l + r) / 2;
  if (il <= m) update(node * 2, l, m, il, ir, x);
  if (m + 1 <= ir) update(node * 2 + 1, m + 1, r, il, ir, x);
  if (a[node * 2].m > a[node * 2 + 1].m)
    a[node].id = a[node * 2].id;
  else
    a[node].id = a[node * 2 + 1].id;
  a[node].m = max(a[node * 2].m, a[node * 2 + 1].m);
  a[node].s = a[node * 2].s + a[node * 2 + 1].s;
}
node query(int nodee, int l, int r, int il, int ir) {
  if (il <= l && r <= ir) return a[nodee];
  node ans = nul;
  int m = (l + r) / 2;
  if (il <= m) {
    node b = query(nodee * 2, l, m, il, ir);
    if (b.m > ans.m) {
      ans.id = b.id;
      ans.m = b.m;
    }
    ans.s = ans.s + b.s;
  }
  if (m + 1 <= ir) {
    node b = query(nodee * 2 + 1, m + 1, r, il, ir);
    if (b.m > ans.m) {
      ans.id = b.id;
      ans.m = b.m;
    }
    ans.s = ans.s + b.s;
  }
  return ans;
}
int main() {
  nul.m = nul.id = nul.s = nul.v = 0;
  cin >> n >> m;
  for (int i = 1, x; i <= n; ++i) {
    cin >> x;
    update(1, 1, n, i, i, x);
  }
  for (int i = 1, t, l, r, x; i <= m; ++i) {
    cin >> t;
    if (t == 1) {
      cin >> l >> r;
      node ans = query(1, 1, n, l, r);
      cout << ans.s << '\n';
    }
    if (t == 2) {
      cin >> l >> r >> x;
      node ans = query(1, 1, n, l, r);
      while (ans.m >= x) {
        update(1, 1, n, ans.id, ans.id, ans.m % x);
        ans = query(1, 1, n, l, r);
      }
    }
    if (t == 3) {
      cin >> l >> x;
      update(1, 1, n, l, l, x);
    }
  }
}