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 = 1e5 + 5;
long long n, m, a[N], T[4 * N], max_value[4 * N];
void build(long long l = 1, long long r = n, long long u = 1) {
  if (l == r) {
    T[u] = a[l];
    max_value[u] = a[l];
    return;
  }
  long long m = (l + r) >> 1;
  build(l, m, u * 2);
  build(m + 1, r, u * 2 + 1);
  max_value[u] = max(max_value[u * 2], max_value[u * 2 + 1]);
  T[u] = T[u * 2] + T[u * 2 + 1];
}
void upd1(long long x, long long ll, long long rr, long long l = 1,
          long long r = n, long long u = 1) {
  if (l > rr || r < ll || max_value[u] < x) return;
  if (l == r) {
    a[l] %= x;
    T[u] = a[l];
    max_value[u] = a[l];
    return;
  }
  long long m = (l + r) >> 1;
  upd1(x, ll, rr, l, m, u * 2);
  upd1(x, ll, rr, m + 1, r, u * 2 + 1);
  T[u] = T[u * 2] + T[u * 2 + 1];
  max_value[u] = max(max_value[u * 2], max_value[u * 2 + 1]);
}
void upd2(long long p, long long x, long long l = 1, long long r = n,
          long long u = 1) {
  if (l > p || r < p) return;
  if (l == r) {
    a[l] = x;
    T[u] = a[l];
    max_value[u] = a[l];
    return;
  }
  long long m = (l + r) >> 1;
  upd2(p, x, l, m, u * 2);
  upd2(p, x, m + 1, r, u * 2 + 1);
  T[u] = T[u * 2] + T[u * 2 + 1];
  max_value[u] = max(max_value[u * 2], max_value[u * 2 + 1]);
}
long long get(long long ll, long long rr, long long l = 1, long long r = n,
              long long u = 1) {
  if (l > rr || r < ll) return 0;
  if (l >= ll && r <= rr) return T[u];
  long long m = (l + r) >> 1;
  return get(ll, rr, l, m, u * 2) + get(ll, rr, m + 1, r, u * 2 + 1);
}
void local_run() {
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  build();
  while (m--) {
    long long v, l, r, x;
    cin >> v >> l >> r;
    if (v == 1)
      cout << get(l, r) << '\n';
    else if (v == 2) {
      cin >> x;
      upd1(x, l, r);
    } else
      upd2(l, r);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  local_run();
}