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 inf = 1e9;
struct node {
  int left, right;
  long long sum = 0;
  long long ma = 0;
  node *child_left, *child_right;
};
node* build(int left, int right, vector<int>& a) {
  if (left > right) {
    return nullptr;
  }
  node* res = new node;
  res->left = left;
  res->right = right;
  if (left == right) {
    res->child_left = res->child_right = nullptr;
    res->sum = a[left];
    res->ma = a[left];
    return res;
  }
  int mid = (left + right) / 2;
  res->child_left = build(left, mid, a);
  res->child_right = build(mid + 1, right, a);
  res->sum = res->child_left->sum + res->child_right->sum;
  res->ma = max(res->child_left->ma, res->child_right->ma);
  return res;
}
void update(node* root, int left, int right, int x) {
  if (right < root->left || left > root->right || x > root->ma) {
    return;
  }
  if (root->left == root->right) {
    root->sum %= x;
    root->ma %= x;
    return;
  }
  update(root->child_left, left, right, x);
  update(root->child_right, left, right, x);
  root->ma = max(root->child_left->ma, root->child_right->ma);
  root->sum = root->child_left->sum + root->child_right->sum;
  return;
}
long long get_sum(node* root, int left, int right) {
  if (right < root->left || left > root->right) {
    return 0;
  }
  if (left <= root->left && root->right <= right) {
    return root->sum;
  }
  long long ansL = get_sum(root->child_left, left, right);
  long long ansR = get_sum(root->child_right, left, right);
  return ansL + ansR;
}
void updateone(node* root, int id, int x) {
  if (root->left > id || root->right < id) return;
  if (root->left == root->right) {
    root->sum = root->ma = x;
    return;
  }
  updateone(root->child_left, id, x);
  updateone(root->child_right, id, x);
  root->ma = max(root->child_left->ma, root->child_right->ma);
  root->sum = root->child_left->sum + root->child_right->sum;
  return;
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  node* root = build(0, n - 1, a);
  for (int i = 0; i < m; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << get_sum(root, l, r) << "\n";
    } else if (type == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      l--;
      r--;
      update(root, l, r, x);
    } else {
      int k, x;
      cin >> k >> x;
      k--;
      updateone(root, k, x);
    }
  }
}