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 + 7;
int a[MAX], n, m;
pair<long long int, long long int> t[4 * MAX];
pair<long long int, long long int> operator+(
    pair<long long int, long long int> a,
    pair<long long int, long long int> b) {
  return {a.first + b.first, max(a.second, b.second)};
}
pair<long long int, long long int> build(int id = 1, int st = 1, int ed = n) {
  if (st == ed) return t[id] = {a[st], a[st]};
  int piv = (st + ed) >> 1;
  return t[id] = build(id << 1, st, piv) + build(id << 1 | 1, piv + 1, ed);
}
pair<long long int, long long int> upd1(int p, long long int x, int id = 1,
                                        int st = 1, int ed = n) {
  if (p > ed || p < st) return t[id];
  if (st == ed) return t[id] = {x, x};
  int piv = (st + ed) >> 1;
  return t[id] = upd1(p, x, id << 1, st, piv) +
                 upd1(p, x, id << 1 | 1, piv + 1, ed);
}
pair<long long int, long long int> upd2(int qx, int qy, int md, int id = 1,
                                        int st = 1, int ed = n) {
  if (qx > ed || qy < st) return t[id];
  if (t[id].second < md) return t[id];
  if (st == ed) return t[id] = {t[id].first % md, t[id].first % md};
  int piv = (st + ed) >> 1;
  return t[id] = upd2(qx, qy, md, id << 1, st, piv) +
                 upd2(qx, qy, md, id << 1 | 1, piv + 1, ed);
}
long long int qry(int qx, int qy, int id = 1, int st = 1, int ed = n) {
  if (qx > ed || qy < st) return 0;
  if (qx <= st && ed <= qy) return t[id].first;
  int piv = (st + ed) >> 1;
  return qry(qx, qy, id << 1, st, piv) + qry(qx, qy, id << 1 | 1, piv + 1, ed);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  build();
  while (m--) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int x, y;
      cin >> x >> y;
      cout << qry(x, y) << '\n';
    } else if (tp == 2) {
      int x, y, md;
      cin >> x >> y >> md;
      upd2(x, y, md);
    } else {
      int p, x;
      cin >> p >> x;
      upd1(p, x);
    }
  }
  return 0;
}