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 = 1e5 + 5;
int n;
long long sum[3 * N], mx[3 * N];
void assign(int i, int x, int u = 1, int s = 0, int e = n - 1) {
  if (i < s or e < i) return;
  if (s == e) {
    sum[u] = mx[u] = x;
    return;
  }
  int v = 2 * u, w = 2 * u + 1, m = (s + e) / 2;
  assign(i, x, v, s, m);
  assign(i, x, w, m + 1, e);
  sum[u] = sum[v] + sum[w];
  mx[u] = max(mx[v], mx[w]);
}
void mod(int l, int r, long long x, int u = 1, int s = 0, int e = n - 1) {
  if (r < s or e < l or mx[u] < x) return;
  if (s == e) {
    mx[u] %= x;
    sum[u] = mx[u];
    return;
  }
  int v = 2 * u, w = 2 * u + 1, m = (s + e) / 2;
  mod(l, r, x, v, s, m);
  mod(l, r, x, w, m + 1, e);
  sum[u] = sum[v] + sum[w];
  mx[u] = max(mx[v], mx[w]);
}
long long rsum(int l, int r, int u = 1, int s = 0, int e = n - 1) {
  if (r < s or e < l) return 0;
  if (l <= s and e <= r) return sum[u];
  int v = 2 * u, w = 2 * u + 1, m = (s + e) / 2;
  return rsum(l, r, v, s, m) + rsum(l, r, w, m + 1, e);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  int m;
  cin >> m;
  int a[n];
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i) {
    assign(i, a[i]);
  }
  while (m--) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << rsum(l, r) << "\n";
    } else if (tp == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      l--, r--;
      mod(l, r, x);
    } else {
      int k, x;
      cin >> k >> x;
      k--;
      assign(k, x);
    }
  }
  return 0;
}