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 + 10;
struct node {
  long long mx;
  long long i;
  long long sum;
  node() {
    mx = 0;
    i = -1;
    sum = 0;
  }
};
node t[4 * maxn];
long long a[maxn];
void upd(long long v, long long vl, long long vr) {
  if (t[vl].mx > t[vr].mx) {
    t[v].i = t[vl].i;
    t[v].mx = t[vl].mx;
  } else {
    t[v].i = t[vr].i;
    t[v].mx = t[vr].mx;
  }
  t[v].sum = t[vl].sum + t[vr].sum;
}
void build_tree(long long v, long long l, long long r) {
  if (l + 1 == r) {
    t[v].mx = a[l];
    t[v].i = l;
    t[v].sum = a[l];
    return;
  }
  long long m = (l + r) / 2;
  long long vl = 2 * v, vr = vl + 1;
  build_tree(vl, l, m);
  build_tree(vr, m, r);
  upd(v, vl, vr);
}
long long st, ed, dt, res, id, sum;
void get_segm(long long v, long long l, long long r) {
  if (l >= ed || r <= st) return;
  if (st <= l && r <= ed) {
    if (res < t[v].mx) {
      res = t[v].mx;
      id = t[v].i;
    }
    sum += t[v].sum;
    return;
  }
  long long m = (l + r) / 2;
  long long vl = 2 * v, vr = vl + 1;
  get_segm(vl, l, m);
  get_segm(vr, m, r);
}
void update(long long v, long long l, long long r) {
  if (l >= ed || r <= st) return;
  if (l + 1 == r) {
    t[v].mx = dt;
    t[v].sum = dt;
    return;
  }
  long long m = (l + r) / 2;
  long long vl = 2 * v, vr = vl + 1;
  update(vl, l, m);
  update(vr, m, r);
  upd(v, vl, vr);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  build_tree(1, 0, n);
  for (long long i = 0; i < m; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l, r;
      cin >> l >> r;
      st = l - 1;
      ed = r;
      sum = 0;
      get_segm(1, 0, n);
      cout << sum << "\n";
    } else if (t == 2) {
      long long l, r, x;
      cin >> l >> r >> x;
      do {
        st = l - 1;
        ed = r;
        res = 0;
        get_segm(1, 0, n);
        if (res >= x) {
          st = id;
          ed = id + 1;
          dt = res % x;
          update(1, 0, n);
        }
      } while (res >= x);
    } else {
      long long k, x;
      cin >> k >> x;
      st = k - 1;
      ed = k;
      dt = x;
      update(1, 0, n);
    }
  }
  return 0;
}