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>
using vc = vector<T>;
using pii = pair<long long, long long>;
void xmax(long long& a, long long b) { a = max(a, b); }
void xmin(long long& a, long long b) { a = min(a, b); }
vc<pii> seg;
long long pow2(long long k) {
  long long kk = 1;
  while (kk < k) kk *= 2;
  return kk;
}
long long query_sum(long long l, long long r) {
  if (r < l) return 0;
  if (r == l) return seg[l].first;
  long long sum = 0;
  if (l % 2 == 1) {
    sum += seg[l].first;
    l++;
  }
  if (r % 2 == 0) {
    sum += seg[r].first;
    r--;
  }
  l /= 2;
  r /= 2;
  return sum + query_sum(l, r);
}
long long query_max(long long l, long long r) {
  if (r < l) return 0;
  if (r == l) return seg[l].second;
  long long maxi = 0;
  if (l % 2 == 1) {
    xmax(maxi, seg[l].second);
    l++;
  }
  if (r % 2 == 0) {
    xmax(maxi, seg[r].second);
    r--;
  }
  l /= 2;
  r /= 2;
  return max(maxi, query_max(l, r));
}
long long nn;
void set_max(long long pos, long long val) {
  if (pos == 0) return;
  if (pos >= nn) {
    seg[pos] = {val, val};
  } else {
    seg[pos] = {seg[2 * pos].first + seg[2 * pos + 1].first,
                max(seg[2 * pos].second, seg[2 * pos + 1].second)};
  }
  set_max(pos / 2, val);
}
void query_dir(long long l, long long x) {
  if (seg[l].second < x) return;
  if (l >= nn) {
    set_max(l, seg[l].first % x);
    return;
  }
  if (seg[2 * l].second >= x) query_dir(2 * l, x);
  if (seg[2 * l + 1].second >= x) query_dir(2 * l + 1, x);
}
void query(long long l, long long r, long long x) {
  if (r < l) return;
  if (query_max(l, r) < x) return;
  if (l == r) {
    query_dir(l, x);
    return;
  }
  if (l % 2 == 1) {
    query_dir(l, x);
    l++;
  }
  if (r % 2 == 0) {
    query_dir(r, x);
    r--;
  }
  query(l / 2, r / 2, x);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vc<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  nn = pow2(n);
  seg = vc<pii>(2 * nn);
  for (long long i = 0; i < n; i++) {
    set_max(i + nn, a[i]);
  }
  long long k = 2;
  for (long long i = 0; i < m; i++) {
    long long mode;
    cin >> mode;
    if (mode == 1) {
      long long l, r;
      cin >> l >> r;
      cout << query_sum(l - 1 + nn, r - 1 + nn) << "\n";
    }
    if (mode == 2) {
      long long l, r, x;
      cin >> l >> r >> x;
      l--;
      r--;
      query(nn + l, nn + r, x);
    }
    if (mode == 3) {
      long long k, x;
      cin >> k >> x;
      set_max(nn + k - 1, x);
    }
  }
}