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 maxn = 1e5;
long long n, m, mod, a[maxn], sum[maxn];
set<pair<long long, long long> > s1[maxn];
inline long long query(long long fi, long long se) {
  long long ret = 0;
  for (long long i = fi; i <= se; i++) {
    if (i % 400 == 0 && i + 400 <= se) {
      ret += sum[i / 400];
      i += 399;
    } else
      ret += a[i];
  }
  return ret;
}
inline void modify(long long ind, long long val) {
  s1[ind / 400].erase(pair<long long, long long>(a[ind], ind));
  sum[ind / 400] -= a[ind];
  a[ind] = val;
  s1[ind / 400].insert(pair<long long, long long>(a[ind], ind));
  sum[ind / 400] += a[ind];
}
inline void modifyseg(long long l, long long r) {
  if (l + 800 > r) {
    for (long long i = l; i <= r; i++)
      if (a[i] >= mod) modify(i, a[i] % mod);
  } else {
    for (long long i = l; i < n && i % 400; i++)
      if (a[i] >= mod) modify(i, a[i] % mod);
    for (long long i = 0; (i + 1) * 400 < n; i++)
      if (l <= (i * 400) && (i * 400 + 399) <= r)
        while (s1[i].rbegin()->first >= mod)
          modify(s1[i].rbegin()->second, s1[i].rbegin()->first % mod);
    for (long long i = r;; i--) {
      if (a[i] >= mod) modify(i, a[i] % mod);
      if (i % 400 == 0) break;
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  mod = 1e10;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s1[i / 400].insert(pair<long long, long long>(a[i], i));
    sum[i / 400] += a[i];
  }
  for (long long type, i = 0; i < m; i++) {
    cin >> type;
    if (type == 1) {
      long long l, r;
      cin >> l >> r;
      cout << query(l - 1, r - 1) << endl;
    }
    if (type == 2) {
      long long l, r, x;
      cin >> l >> r >> x;
      mod = x;
      modifyseg(l - 1, r - 1);
    }
    if (type == 3) {
      long long ind, val;
      cin >> ind >> val;
      ind--;
      modify(ind, val);
    }
  }
}