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 = 100005;
int n, q, ql;
long long a[N];
long long mod, qr;
struct node {
  pair<long long, int> Max;
  long long S = 0;
};
node seg[4 * N];
vector<node*> v;
void read() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
}
void build(int l = 0, int r = n, int id = 0) {
  if (r - l < 2) {
    seg[id].S = a[l];
    seg[id].Max = {a[l], l};
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, 2 * id + 1);
  build(mid, r, 2 * id + 2);
  seg[id].S = seg[2 * id + 1].S + seg[2 * id + 2].S;
  seg[id].Max = max(seg[2 * id + 1].Max, seg[2 * id + 2].Max);
}
long long getsum(int l = 0, int r = n, int id = 0) {
  if (qr <= l || r <= ql) return 0LL;
  if (ql <= l && r <= qr) return seg[id].S;
  int mid = (l + r) / 2;
  return getsum(l, mid, 2 * id + 1) + getsum(mid, r, 2 * id + 2);
}
void assign(int l = 0, int r = n, int id = 0) {
  if (ql < l || r <= ql) return;
  if (r - l < 2) {
    seg[id].S = qr;
    seg[id].Max = {qr, ql};
    return;
  }
  int mid = (l + r) / 2;
  assign(l, mid, 2 * id + 1);
  assign(mid, r, 2 * id + 2);
  seg[id].S = seg[2 * id + 1].S + seg[2 * id + 2].S;
  seg[id].Max = max(seg[2 * id + 1].Max, seg[2 * id + 2].Max);
}
void findR(int l = 0, int r = n, int id = 0) {
  if (qr <= l || r <= ql) return;
  if (ql <= l && r <= qr) {
    v.push_back(&seg[id]);
    return;
  }
  int mid = (l + r) / 2;
  findR(l, mid, 2 * id + 1);
  findR(mid, r, 2 * id + 2);
}
void print() {
  for (int i = 0; i < 4 * n; i++) {
    cout << "seg[" << i << "].S   = " << seg[i].S << '\n';
    cout << "seg[" << i << "].Max = " << seg[i].Max.first << '\n';
    cout << "\n--------------------------------\n";
  }
}
void query() {
  while (q--) {
    int act;
    cin >> act >> ql >> qr;
    ql--;
    if (act == 1) {
      cout << getsum() << '\n';
    }
    if (act == 2) {
      cin >> mod;
      v.clear();
      findR();
      for (auto u : v) {
        auto it = u->Max;
        while (it.first >= mod) {
          ql = it.second;
          qr = ((it.first) % mod);
          assign();
          it = u->Max;
        }
      }
    }
    if (act == 3) {
      assign();
    }
  }
}
int main() {
  read();
  build();
  query();
  return 0;
}