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;
int n, m;
const int size = 311111;
int ans[size], idx[size], a[size];
long long sum[size];
void adp(int p) {
  int x = p * 2;
  if (ans[x] < ans[x + 1]) x++;
  ans[p] = ans[x];
  idx[p] = idx[x];
  sum[p] = sum[x] + sum[x ^ 1];
}
void init(int p, int l, int r) {
  if (l == r) {
    sum[p] = ans[p] = a[l];
    idx[p] = l;
    return;
  }
  int mid = (l + r) / 2;
  init(p * 2, l, mid);
  init(p * 2 + 1, mid + 1, r);
  adp(p);
}
void change(int p, int l, int r, int x, int num) {
  if (l == r) {
    sum[p] = ans[p] = num;
    return;
  }
  int mid = (l + r) / 2;
  if (mid >= x)
    change(p * 2, l, mid, x, num);
  else
    change(p * 2 + 1, mid + 1, r, x, num);
  adp(p);
}
long long get(int p, int l, int r, int L, int R) {
  if (L <= l && R >= r) return sum[p];
  if (L > r || R < l) return 0;
  int mid = (l + r) / 2;
  return get(p * 2, l, mid, L, R) + get(p * 2 + 1, mid + 1, r, L, R);
}
void pause() {
  for (int i = 1; i <= n; i++) cout << get(1, 1, n, i, i) << ' ';
  cout << endl;
  int a;
  cin >> a;
}
void deal(int p, int l, int r, int L, int R, int x) {
  if (L <= l && R >= r) {
    while (ans[p] >= x) {
      change(1, 1, n, idx[p], ans[p] % x);
    }
    return;
  }
  if (L > r || R < l) return;
  int mid = (l + r) / 2;
  deal(p * 2, l, mid, L, R, x);
  deal(p * 2 + 1, mid + 1, r, L, R, x);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  init(1, 1, n);
  while (m--) {
    int t, l, r, x;
    cin >> t;
    if (t == 1) {
      cin >> l >> r;
      cout << get(1, 1, n, l, r) << endl;
    } else if (t == 2) {
      cin >> l >> r >> x;
      deal(1, 1, n, l, r, x);
    } else if (t == 3) {
      cin >> l >> x;
      change(1, 1, n, l, x);
    }
  }
}