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;
struct node {
  long long sum, mx;
  node() {}
  node(long long v) : sum(v), mx(v) {}
  node(long long sum, long long mx) : sum(sum), mx(mx) {}
};
node NEUT(0);
node oper(node x, node y) { return node(x.sum + y.sum, max(x.mx, y.mx)); }
struct STree {
  vector<node> st;
  int n;
  STree(int n) : st(4 * n + 5, NEUT), n(n) {}
  void upd(int k, int s, int e, int p, int v) {
    if (s + 1 == e) {
      st[k] = node(v);
      return;
    }
    int m = (s + e) / 2;
    if (p < m)
      upd(2 * k, s, m, p, v);
    else
      upd(2 * k + 1, m, e, p, v);
    st[k] = oper(st[2 * k], st[2 * k + 1]);
  }
  node query(int k, int s, int e, int a, int b) {
    if (s >= b || e <= a) return NEUT;
    if (s >= a && e <= b) return st[k];
    int m = (s + e) / 2;
    return oper(query(2 * k, s, m, a, b), query(2 * k + 1, m, e, a, b));
  }
  int get(int k, int s, int e, int pos, int v) {
    if (e <= pos) return n;
    if (s >= pos) {
      if (st[k].mx < v) return n;
      if (s + 1 == e) return s;
    }
    int m = (s + e) / 2;
    int ans = get(2 * k, s, m, pos, v);
    if (ans == n) ans = get(2 * k + 1, m, e, pos, v);
    return ans;
  }
  void upd(int p, int v) { upd(1, 0, n, p, v); }
  int get(int pos, int v) { return get(1, 0, n, pos, v); }
  node query(int a, int b) { return query(1, 0, n, a, b); }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  STree st(n);
  vector<int> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i], st.upd(i, a[i]);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      cout << st.query(l - 1, r).sum << "\n";
    }
    if (t == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      l--;
      r--;
      while (l <= r) {
        st.upd(l, a[l] % x);
        a[l] %= x;
        l = st.get(l + 1, x);
      }
    }
    if (t == 3) {
      int k, x;
      cin >> k >> x;
      k--;
      a[k] = x;
      st.upd(k, x);
    }
  }
  return 0;
}