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 MAX = 1e5 + 1;
struct node {
  long long mx;
  long long sum;
} seg[4 * MAX];
long long a[MAX];
void build(int i, int st, int en) {
  if (st == en) {
    seg[i].mx = seg[i].sum = a[st];
    return;
  }
  int mid = (st + en) >> 1, lnode = i << 1, rnode = (i << 1) | 1;
  build(lnode, st, mid);
  build(rnode, mid + 1, en);
  seg[i].sum = seg[lnode].sum + seg[rnode].sum;
  seg[i].mx = max(seg[lnode].mx, seg[rnode].mx);
}
void update1(int i, int st, int en, int l, int r, int mod) {
  if (l > en || r < st) {
    return;
  }
  if (st == en) {
    a[st] %= mod;
    seg[i].mx = seg[i].sum = seg[i].sum % mod;
    return;
  }
  int mid = (st + en) >> 1, lnode = i << 1, rnode = (i << 1) | 1;
  if (seg[lnode].mx >= mod) {
    update1(lnode, st, mid, l, r, mod);
  }
  if (seg[rnode].mx >= mod) {
    update1(rnode, mid + 1, en, l, r, mod);
  }
  seg[i].sum = seg[lnode].sum + seg[rnode].sum;
  seg[i].mx = max(seg[lnode].mx, seg[rnode].mx);
}
void update2(int i, int st, int en, int idx, int val) {
  if (st == en) {
    a[st] = val;
    seg[i].sum = seg[i].mx = val;
    return;
  }
  int mid = (st + en) >> 1, lnode = i << 1, rnode = (i << 1) | 1;
  if (idx > mid) {
    update2(rnode, mid + 1, en, idx, val);
  } else {
    update2(lnode, st, mid, idx, val);
  }
  seg[i].sum = seg[lnode].sum + seg[rnode].sum;
  seg[i].mx = max(seg[lnode].mx, seg[rnode].mx);
}
long long query(int i, int st, int en, int l, int r) {
  if (l > en || r < st) {
    return 0;
  }
  if (l <= st && en <= r) {
    return seg[i].sum;
  }
  int mid = (st + en) >> 1, lnode = i << 1, rnode = (i << 1) | 1;
  return query(lnode, st, mid, l, r) + query(rnode, mid + 1, en, l, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, 1, n);
  int l, r, x, k, q;
  while (m--) {
    cin >> q;
    if (q == 1) {
      cin >> l >> r;
      cout << query(1, 1, n, l, r) << "\n";
    } else if (q == 2) {
      cin >> l >> r >> x;
      update1(1, 1, n, l, r, x);
    } else {
      cin >> k >> x;
      update2(1, 1, n, k, x);
    }
  }
  return 0;
}