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 main() {
  long long n, m;
  cin >> n >> m;
  long long a[100005] = {0};
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long SGT[100005 * 4][2];
  memset(SGT, 0, sizeof SGT);
  function<void(long long, long long, long long)> build =
      [&](long long id, long long s, long long e) {
        if (s == e) {
          SGT[id][0] = a[s];
          SGT[id][1] = a[s];
          return;
        }
        long long mid = (s + e) / 2;
        build(id + id, s, mid);
        build(id + id + 1, mid + 1, e);
        SGT[id][0] = max(SGT[id + id][0], SGT[id + id + 1][0]);
        SGT[id][1] = SGT[id + id][1] + SGT[id + id + 1][1];
      };
  build(1, 1, n);
  function<void(long long, long long, long long, long long, long long,
                long long)>
      update_mod = [&](long long id, long long s, long long e, long long l,
                       long long r, long long md) {
        if (s > r || l > e) return;
        if (s == e) {
          SGT[id][0] = a[s] % md;
          SGT[id][1] = a[s] % md;
          a[s] %= md;
          return;
        }
        if (l <= s && e <= r && SGT[id][0] < md) return;
        long long mid = (s + e) / 2;
        update_mod(id + id, s, mid, l, r, md);
        update_mod(id + id + 1, mid + 1, e, l, r, md);
        SGT[id][0] = max(SGT[id + id][0], SGT[id + id + 1][0]);
        SGT[id][1] = SGT[id + id + 1][1] + SGT[id + id][1];
      };
  function<void(long long, long long, long long, long long, long long)>
      update_ind = [&](long long id, long long s, long long e, long long ind,
                       long long val) {
        if (s == e) {
          SGT[id][0] = val;
          SGT[id][1] = val;
          a[s] = val;
          return;
        }
        long long mid = (s + e) / 2;
        if (ind <= mid)
          update_ind(id + id, s, mid, ind, val);
        else
          update_ind(id + id + 1, mid + 1, e, ind, val);
        SGT[id][1] = SGT[id + id + 1][1] + SGT[id + id][1];
        SGT[id][0] = max(SGT[id + id + 1][0], SGT[id + id][0]);
      };
  function<long long(long long, long long, long long, long long, long long)>
      query = [&](long long id, long long s, long long e, long long l,
                  long long r) {
        if (l > e || s > r) return 0ll;
        if (l <= s && e <= r) return SGT[id][1];
        long long mid = (s + e) / 2;
        return query(id + id, s, mid, l, r) +
               query(id + id + 1, mid + 1, e, l, r);
      };
  for (long long i = 1; i <= m; i++) {
    long long ty, l, r;
    cin >> ty;
    if (ty == 1) {
      cin >> l >> r;
      cout << query(1, 1, n, l, r) << endl;
    } else if (ty == 2) {
      long long m;
      cin >> l >> r >> m;
      update_mod(1, 1, n, l, r, m);
    } else {
      cin >> l >> r;
      update_ind(1, 1, n, l, r);
    }
  }
  return 0;
}