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 N = (int)1e5 + 11;
int n, m, l, r, ind[N * 4];
long long mn[N * 4], a[N], x, tp, sum[N * 4];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    sum[v] = mn[v] = a[tl];
    ind[v] = tl;
  } else {
    int m = (tl + tr) / 2;
    build(v + v, tl, m);
    build(v + v + 1, m + 1, tr);
    sum[v] = sum[v + v] + sum[v + v + 1];
    mn[v] = max(mn[v + v], mn[v + v + 1]);
    if (mn[v + v] > mn[v + v + 1])
      ind[v] = ind[v + v];
    else
      ind[v] = ind[v + v + 1];
  }
}
void upd(int v, int tl, int tr, int pos, long long val) {
  if (tl == tr)
    sum[v] = mn[v] = val;
  else {
    int m = (tl + tr) / 2;
    if (pos <= m)
      upd(v + v, tl, m, pos, val);
    else
      upd(v + v + 1, m + 1, tr, pos, val);
    sum[v] = sum[v + v] + sum[v + v + 1];
    mn[v] = max(mn[v + v], mn[v + v + 1]);
    if (mn[v + v] > mn[v + v + 1])
      ind[v] = ind[v + v];
    else
      ind[v] = ind[v + v + 1];
  }
}
long long get_sum(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (l == tl && r == tr) return sum[v];
  int m = (tl + tr) / 2;
  return get_sum(v + v, tl, m, l, min(r, m)) +
         get_sum(v + v + 1, m + 1, tr, max(l, m + 1), r);
}
pair<long long, int> get_mn(int v, int tl, int tr, int l, int r) {
  if (l > r) return make_pair((int)2e9 * -1, 0);
  if (l == tl && r == tr) return make_pair(mn[v], ind[v]);
  int m = (tl + tr) / 2;
  pair<int, int> L, R;
  L = get_mn(v + v, tl, m, l, min(r, m));
  R = get_mn(v + v + 1, m + 1, tr, max(l, m + 1), r);
  if (L.first > R.first) return L;
  return R;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  for (int i = 0; i < m; i++) {
    cin >> tp;
    if (tp == 1) {
      cin >> l >> r;
      cout << get_sum(1, 1, n, l, r) << endl;
    } else if (tp == 2) {
      cin >> l >> r >> x;
      while (true) {
        pair<long long, int> cur = get_mn(1, 1, n, l, r);
        if (cur.first < x) break;
        upd(1, 1, n, cur.second, cur.first % x);
      }
    } else {
      cin >> l >> x;
      upd(1, 1, n, l, x);
    }
  }
  return 0;
}