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;
long long arr[100005];
pair<long long, long long> tree[300005];
long long n, m;
void update(long long i, long long l, long long r, long long pos,
            long long val) {
  if (l == r) {
    tree[i] = {val, val};
    return;
  }
  long long mid = (l + r) / 2;
  if (pos <= mid)
    update(i * 2, l, mid, pos, val);
  else
    update(i * 2 + 1, mid + 1, r, pos, val);
  tree[i] = {tree[i * 2].first + tree[i * 2 + 1].first,
             max(tree[i * 2].second, tree[i * 2 + 1].second)};
}
long long query(long long i, long long l, long long r, long long l1,
                long long r1) {
  if (l1 <= l && r <= r1) return tree[i].first;
  if (r < l1 || l > r1) return 0;
  long long mid = (l + r) / 2;
  return query(i * 2, l, mid, l1, r1) + query(i * 2 + 1, mid + 1, r, l1, r1);
}
void updatemod(long long i, long long l, long long r, long long l1,
               long long r1, long long mod) {
  if (tree[i].second < mod) return;
  if (l == r) {
    tree[i] = {tree[i].first % mod, tree[i].second % mod};
    return;
  }
  long long mid = (l + r) / 2;
  if (r1 <= mid)
    updatemod(i * 2, l, mid, l1, r1, mod);
  else if (l1 > mid)
    updatemod(i * 2 + 1, mid + 1, r, l1, r1, mod);
  else {
    updatemod(i * 2, l, mid, l1, r1, mod);
    updatemod(i * 2 + 1, mid + 1, r, l1, r1, mod);
  }
  tree[i] = {tree[i * 2].first + tree[i * 2 + 1].first,
             max(tree[i * 2].second, tree[i * 2 + 1].second)};
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long m;
  cin >> n >> m;
  long long arr[n + 2];
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  for (long long i = 1; i <= n; i++) update(1, 1, n, i, arr[i]);
  while (m--) {
    long long q;
    cin >> q;
    if (q == 1) {
      long long l, r;
      cin >> l >> r;
      cout << query(1, 1, n, l, r) << '\n';
    } else if (q == 3) {
      long long pos, val;
      cin >> pos >> val;
      arr[pos] = val;
      update(1, 1, n, pos, arr[pos]);
    } else {
      long long l, r, x;
      cin >> l >> r >> x;
      updatemod(1, 1, n, l, r, x);
    }
  }
}