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 int Maxn = 555555;
const long long INF = -1e18;
long long n, q, a[Maxn], sum[Maxn];
pair<long long, long long> mx[Maxn];
void updatemx(long long st, long long en, long long tag, long long pos,
              long long val) {
  if (st + 1 == en) {
    mx[tag].first = val, mx[tag].second = pos;
    return;
  }
  long long mid = (st + en) >> 1;
  if (pos < mid)
    updatemx(st, mid, tag * 2, pos, val);
  else
    updatemx(mid, en, tag * 2 + 1, pos, val);
  mx[tag] = max(mx[tag * 2], mx[tag * 2 + 1]);
}
pair<long long, long long> getmx(long long st, long long en, long long tag,
                                 long long lo, long long hi) {
  if (en <= lo || hi <= st) return make_pair(INF, (long long)0);
  if (lo <= st && en <= hi) return mx[tag];
  long long mid = (st + en) >> 1;
  return max(getmx(st, mid, tag * 2, lo, hi),
             getmx(mid, en, tag * 2 + 1, lo, hi));
}
void updatesum(long long st, long long en, long long tag, long long pos,
               long long val) {
  if (st + 1 == en) {
    sum[tag] = val;
    return;
  }
  long long mid = (st + en) >> 1;
  if (pos < mid)
    updatesum(st, mid, tag * 2, pos, val);
  else
    updatesum(mid, en, tag * 2 + 1, pos, val);
  sum[tag] = sum[tag * 2] + sum[tag * 2 + 1];
}
long long getsum(long long st, long long en, long long tag, long long lo,
                 long long hi) {
  if (en <= lo || hi <= st) return 0;
  if (lo <= st && en <= hi) return sum[tag];
  long long mid = (st + en) >> 1;
  return getsum(st, mid, tag * 2, lo, hi) +
         getsum(mid, en, tag * 2 + 1, lo, hi);
}
int main(int argc, char** argv) {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    updatemx(1, n + 1, 1, i, a[i]);
    updatesum(1, n + 1, 1, i, a[i]);
  }
  for (int i = 1; i <= q; i++) {
    long long type;
    cin >> type;
    long long l, r, x, pos, val;
    pair<long long, long long> p;
    switch (type) {
      case (1):
        cin >> l >> r;
        cout << getsum(1, n + 1, 1, l, ++r) << endl;
        break;
      case (2):
        cin >> l >> r >> x;
        r++;
        p = getmx(1, n + 1, 1, l, r);
        while (p.first >= x) {
          updatemx(1, n + 1, 1, p.second, p.first % x);
          updatesum(1, n + 1, 1, p.second, p.first % x);
          p = getmx(1, n + 1, 1, l, r);
        }
        break;
      case (3):
        cin >> pos >> val;
        updatemx(1, n + 1, 1, pos, val);
        updatesum(1, n + 1, 1, pos, val);
        break;
    }
  }
  return 0;
}