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;
template <typename T>
inline bool smin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool smax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const long long N = (long long)1e5 + 10;
long long n, ar[N], q, sum[N << 2], mx[N << 2], ind[N << 2];
void change(long long pos, long long val, long long id = 1, long long l = 0,
            long long r = n) {
  if (r - l < 2) {
    sum[id] = mx[id] = val;
    ind[id] = pos;
    return;
  }
  long long mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  if (pos < mid)
    change(pos, val, lc, l, mid);
  else
    change(pos, val, rc, mid, r);
  sum[id] = sum[lc] + sum[rc];
  mx[id] = max(mx[lc], mx[rc]);
  ind[id] = ind[rc];
  if (mx[id] == mx[lc]) ind[id] = ind[lc];
}
long long get_max(long long x, long long y, long long id = 1, long long l = 0,
                  long long r = n) {
  if (r <= x || l >= y) return -1;
  if (l >= x && r <= y) return ind[id];
  long long mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  long long pos0 = get_max(x, y, lc, l, mid);
  long long pos1 = get_max(x, y, rc, mid, r);
  if (pos0 == -1) return pos1;
  if (pos1 == -1) return pos0;
  if (ar[pos0] >= ar[pos1]) return pos0;
  return pos1;
}
long long get_sum(long long x, long long y, long long id = 1, long long l = 0,
                  long long r = n) {
  if (r <= x || l >= y) return 0;
  if (l >= x && r <= y) return sum[id];
  long long mid = (r + l) >> 1, lc = id << 1, rc = lc | 1;
  return get_sum(x, y, lc, l, mid) + get_sum(x, y, rc, mid, r);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long j = 0; j < n; ++j) cin >> ar[j];
  for (long long j = 0; j < n; ++j) change(j, ar[j]);
  while (q--) {
    long long typ;
    cin >> typ;
    if (typ == 1) {
      long long bl, br;
      cin >> bl >> br;
      --bl;
      cout << get_sum(bl, br) << '\n';
    } else if (typ == 2) {
      long long bl, br, x;
      cin >> bl >> br >> x;
      --bl;
      while (true) {
        long long pos = get_max(bl, br);
        if (ar[pos] < x) break;
        ar[pos] %= x;
        change(pos, ar[pos]);
      }
    } else {
      long long pos, val;
      cin >> pos >> val;
      ar[--pos] = val;
      change(pos, val);
    }
  }
}