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 int arr[100003];
struct node {
  long long int max;
  long long int sum;
} t[4 * 100003];
void build(long long int node, long long int start, long long int end) {
  if (start == end) {
    t[node].sum = arr[start];
    t[node].max = arr[start];
  } else {
    long long int mid = (start + end) / 2;
    build(2 * node, start, mid);
    build(2 * node + 1, mid + 1, end);
    t[node].max = max(t[2 * node].max, t[2 * node + 1].max);
    t[node].sum = t[2 * node].sum + t[2 * node + 1].sum;
  }
}
void updateRange(long long int node, long long int start, long long int end,
                 long long int l, long long int r, long long int val) {
  if (t[node].max < val) return;
  if (start == end) {
    t[node].sum = t[node].sum % val;
    t[node].max = t[node].max % val;
    return;
  }
  long long int mid = (start + end) / 2;
  if (l <= mid) updateRange(2 * node, start, mid, l, r, val);
  if (r > mid) updateRange(2 * node + 1, mid + 1, end, l, r, val);
  t[node].max = max(t[2 * node].max, t[2 * node + 1].max);
  t[node].sum = t[2 * node].sum + t[2 * node + 1].sum;
  return;
}
void update(long long int node, long long int start, long long int end,
            long long int idx, long long int val) {
  if (start == end) {
    t[node].max = val;
    t[node].sum = val;
  } else {
    long long int mid = (start + end) / 2;
    if (idx <= mid)
      update(2 * node, start, mid, idx, val);
    else
      update(2 * node + 1, mid + 1, end, idx, val);
    t[node].max = max(t[2 * node].max, t[2 * node + 1].max);
    t[node].sum = t[2 * node].sum + t[2 * node + 1].sum;
  }
}
long long int query(long long int node, long long int start, long long int end,
                    long long int l, long long int r) {
  if (start >= l and end <= r) return t[node].sum;
  long long int mid = (start + end) / 2;
  long long int ans = 0;
  if (l <= mid) ans += query(2 * node, start, mid, l, r);
  if (r > mid) ans += query(2 * node + 1, mid + 1, end, l, r);
  return ans;
}
signed main() {
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  build(1, 1, n);
  while (m--) {
    long long int type;
    cin >> type;
    if (type == 1) {
      long long int l, r;
      cin >> l >> r;
      cout << query(1, 1, n, l, r) << endl;
    } else if (type == 2) {
      long long int l, r, mod;
      cin >> l >> r >> mod;
      updateRange(1, 1, n, l, r, mod);
    } else {
      long long int pos, val;
      cin >> pos >> val;
      update(1, 1, n, pos, val);
    }
  }
  return 0;
}