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 long long N = 2e5 + 30, Mod = 1e9 + 7;
const long long SQ = 330;
const long long INF = 1e17;
long long a[N], n, q;
long long sum[N << 2], mx[N << 2];
void Change(int x, int val, int id = 1, int b = 0, int e = n) {
  if (e - b == 1) {
    mx[id] = a[x] = val;
    sum[id] = val;
    return;
  }
  if (x < ((b + e) >> 1))
    Change(x, val, id << 1, b, ((b + e) >> 1));
  else
    Change(x, val, id << 1 | 1, ((b + e) >> 1), e);
  sum[id] = sum[id << 1] + sum[id << 1 | 1];
  mx[id] = max(mx[id << 1], mx[id << 1 | 1]);
}
long long Get(int l, int r, int id = 1, int b = 0, int e = n) {
  if (r <= b || e <= l) return 0;
  if (l <= b && e <= r) return sum[id];
  return Get(l, r, id << 1, b, ((b + e) >> 1)) +
         Get(l, r, id << 1 | 1, ((b + e) >> 1), e);
}
void Set(int l, int r, int m, int id = 1, int b = 0, int e = n) {
  if (mx[id] < m) return;
  if (r <= b || e <= l) return;
  if (e - b == 1) {
    sum[id] = mx[id] = mx[id] % m;
    return;
  }
  Set(l, r, m, id << 1, b, ((b + e) >> 1));
  Set(l, r, m, id << 1 | 1, ((b + e) >> 1), e);
  sum[id] = sum[id << 1] + sum[id << 1 | 1];
  mx[id] = max(mx[id << 1], mx[id << 1 | 1]);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) Change(i, a[i]);
  while (q--) {
    long long t;
    cin >> t;
    switch (t) {
      case 1:
        int l, r;
        cin >> l >> r;
        l--;
        cout << Get(l, r) << '\n';
        break;
      case 2:
        int x;
        cin >> l >> r >> x;
        l--;
        Set(l, r, x);
        break;
      case 3:
        int k;
        cin >> k >> x;
        Change(k - 1, x);
    }
  }
  return (0);
}