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 long long int maxn = 1e5 + 5, M = 1e6 + 5, bghs = 1e9 + 7, inf = 1e18,
                    Z = maxn / 2;
long long int n, m, a[maxn];
long long int seg[3 * maxn], lazy[3 * maxn], seg2[maxn * 3];
void build(long long int l, long long int r, long long int id) {
  if (r - l == 1) {
    seg2[id] = seg[id] = a[l];
    return;
  }
  long long int mid = (l + r) >> 1;
  build(l, mid, 2 * id + 1);
  build(mid, r, 2 * id + 2);
  seg[id] = seg[id * 2 + 1] + seg[id * 2 + 2];
  seg2[id] = max(seg2[id * 2 + 1], seg2[id * 2 + 2]);
}
long long int get(long long int s, long long int e, long long int l,
                  long long int r, long long int id) {
  if (s <= l && r <= e) return (seg[id]);
  if (e <= l || r <= s) return 0;
  long long int mid = (l + r) >> 1;
  return get(s, e, l, mid, 2 * id + 1) + get(s, e, mid, r, 2 * id + 2);
}
void upd1(long long int x, long long int y, long long int l, long long int r,
          long long int id) {
  if (r - l == 1) {
    seg2[id] = seg[id] = y;
    return;
  }
  long long int mid = (r + l) >> 1;
  if (x < mid)
    upd1(x, y, l, mid, 2 * id + 1);
  else
    upd1(x, y, mid, r, 2 * id + 2);
  seg[id] = seg[2 * id + 2] + seg[2 * id + 1];
  seg2[id] = max(seg2[id * 2 + 1], seg2[id * 2 + 2]);
}
void upd2(long long int s, long long int e, long long int val, long long int l,
          long long int r, long long int id) {
  if (e <= l || r <= s || seg2[id] < val) return;
  if (r - l == 1) {
    seg[id] %= val;
    seg2[id] %= val;
    return;
  }
  long long int mid = (l + r) >> 1;
  upd2(s, e, val, l, mid, 2 * id + 1);
  upd2(s, e, val, mid, r, 2 * id + 2);
  seg[id] = seg[id * 2 + 1] + seg[id * 2 + 2];
  seg2[id] = max(seg2[id * 2 + 1], seg2[id * 2 + 2]);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  build(0, n, 0);
  while (m--) {
    long long int t;
    cin >> t;
    if (t == 1) {
      long long int l, r;
      cin >> l >> r;
      l--;
      cout << get(l, r, 0, n, 0) << '\n';
    }
    if (t == 2) {
      long long int l, r, x;
      cin >> l >> r >> x;
      l--;
      upd2(l, r, x, 0, n, 0);
    }
    if (t == 3) {
      long long int y, x;
      cin >> x >> y;
      x--;
      upd1(x, y, 0, n, 0);
    }
  }
}