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;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
  if (b == 0) return 1;
  long long x = fp(a, b / 2);
  x = (x * x) % mod;
  if (b & 1) x = (x * a) % mod;
  return x;
}
const long long N = 1e5 + 5;
long long a[N], n;
struct node {
  long long mx, sm;
};
node merge(node a, node b) {
  node c;
  c.mx = max(a.mx, b.mx);
  c.sm = a.sm + b.sm;
  return c;
}
vector<node> bit(4 * N);
void build(long long id, long long l, long long r) {
  if (l == r) {
    bit[id].mx = a[l];
    bit[id].sm = a[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * id, l, mid);
  build(2 * id + 1, mid + 1, r);
  bit[id] = merge(bit[2 * id], bit[2 * id + 1]);
}
long long get(long long id, long long l, long long r, long long L,
              long long R) {
  if (R < l || r < L) {
    return 0;
  }
  if (L <= l && r <= R) {
    return bit[id].sm;
  }
  long long mid = (l + r) / 2;
  return get(2 * id, l, mid, L, R) + get(2 * id + 1, mid + 1, r, L, R);
}
void upd3(long long id, long long l, long long r, long long L, long long R,
          long long v) {
  if (R < l || r < L) return;
  if (L <= l && r <= R) {
    bit[id].mx = v;
    bit[id].sm = v;
    return;
  }
  long long mid = (l + r) / 2;
  upd3(2 * id, l, mid, L, R, v);
  upd3(2 * id + 1, mid + 1, r, L, R, v);
  bit[id] = merge(bit[2 * id], bit[2 * id + 1]);
}
void upd2(long long id, long long l, long long r, long long L, long long R,
          long long v) {
  if (bit[id].mx < v || R < l || r < L) return;
  if (L <= l && r <= R && l == r) {
    bit[id].mx %= v;
    bit[id].sm %= v;
    return;
  }
  long long mid = (l + r) / 2;
  upd2(2 * id, l, mid, L, R, v);
  upd2(2 * id + 1, mid + 1, r, L, R, v);
  bit[id] = merge(bit[2 * id], bit[2 * id + 1]);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(0));
  long long m, typ, l, r, x, k;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, 1, n);
  while (m--) {
    cin >> typ;
    if (typ == 1) {
      cin >> l >> r;
      cout << get(1, 1, n, l, r) << "\n";
    } else if (typ == 2) {
      cin >> l >> r >> x;
      upd2(1, 1, n, l, r, x);
    } else {
      cin >> k >> x;
      upd3(1, 1, n, k, k, x);
    }
  }
}