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;
ifstream fin("in.in");
ofstream fout("out.out");
const int N = 1e5 + 10;
int n, m, a[N];
pair<long long, int> node[(N << 2) + 10];
long long getSum(int s, int e, int tag, int l, int r) {
  if (l >= r || s >= r || e <= l) return 0;
  if (s >= l && e <= r) return node[tag].first;
  int mid = (s + e) >> 1;
  return getSum(s, mid, tag * 2, l, r) + getSum(mid, e, tag * 2 + 1, l, r);
}
int getMax(int s, int e, int tag, int l, int r) {
  if (l >= r || s >= r || e <= l) return n;
  if (s >= l && e <= r) return node[tag].second;
  int mid = (s + e) >> 1;
  int mx1 = getMax(s, mid, tag * 2, l, r),
      mx2 = getMax(mid, e, tag * 2 + 1, l, r);
  if (a[mx1] > a[mx2]) return mx1;
  return mx2;
}
void update(int s, int e, int tag, int pos) {
  if (s + 1 == e) {
    node[tag].first = a[pos];
    return;
  }
  int mid = (s + e) >> 1;
  if (pos < mid)
    update(s, mid, tag * 2, pos);
  else
    update(mid, e, tag * 2 + 1, pos);
  node[tag].first = node[tag * 2].first + node[tag * 2 + 1].first;
  if (a[node[tag * 2].second] > a[node[tag * 2 + 1].second])
    node[tag].second = node[tag * 2].second;
  else
    node[tag].second = node[tag * 2 + 1].second;
}
void buildTree(int s, int e, int tag) {
  if (s + 1 == e) {
    node[tag].first = a[s];
    node[tag].second = s;
    return;
  }
  int mid = (s + e) >> 1;
  buildTree(s, mid, tag * 2);
  buildTree(mid, e, tag * 2 + 1);
  node[tag].first = node[tag * 2].first + node[tag * 2 + 1].first;
  if (a[node[tag * 2].second] > a[node[tag * 2 + 1].second])
    node[tag].second = node[tag * 2].second;
  else
    node[tag].second = node[tag * 2 + 1].second;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  buildTree(0, n, 1);
  while (m) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      cout << getSum(0, n, 1, l - 1, r) << endl;
      m--;
      continue;
    }
    if (t == 3) {
      int k, x;
      cin >> k >> x;
      a[--k] = x;
      update(0, n, 1, k);
      m--;
      continue;
    }
    int l, r, x;
    cin >> l >> r >> x;
    int k = getMax(0, n, 1, --l, r);
    while (a[k] >= x) {
      a[k] %= x;
      update(0, n, 1, k);
      k = getMax(0, n, 1, l, r);
    }
    m--;
  }
  return 0;
}