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 MX = 3e5;
int N, M, k;
int arr[MX];
pair<long long, int> st[MX * 4];
void build(int L, int R, int p) {
  if (L == R) {
    st[p] = make_pair(arr[L], arr[L]);
  } else {
    int mid = (L + R) >> 1;
    build(L, mid, p << 1);
    build(mid + 1, R, p << 1 | 1);
    st[p].first = st[p << 1].first + st[p << 1 | 1].first;
    st[p].second = max(st[p << 1].second, st[p << 1 | 1].second);
  }
}
void update(int L, int R, int p, int i, int j, int val, int s) {
  if (L > R || L > j || R < i) return;
  if (L >= i && R <= j) {
    if (s) {
      st[p] = make_pair(val, val);
      return;
    }
    if (st[p].second < val) return;
    if (L == R) {
      st[p] = make_pair(st[p].first % val, st[p].second % val);
      return;
    }
  }
  int mid = (R + L) >> 1;
  update(L, mid, p << 1, i, j, val, s);
  update(mid + 1, R, p << 1 | 1, i, j, val, s);
  st[p].first = st[p << 1].first + st[p << 1 | 1].first;
  st[p].second = max(st[p << 1].second, st[p << 1 | 1].second);
}
long long query(int L, int R, int p, int i, int j) {
  if (L > j || R < i) return 0;
  if (L >= i && R <= j) return st[p].first;
  int mid = (L + R) >> 1;
  return query(L, mid, p << 1, i, j) + query(mid + 1, R, p << 1 | 1, i, j);
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; i++) scanf("%d", &arr[i]);
  build(0, N - 1, 1);
  vector<long long> an;
  while (M--) {
    int t, l, r, val;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &l, &r);
      an.push_back(query(0, N - 1, 1, l - 1, r - 1));
    } else if (t == 2) {
      int val;
      scanf("%d%d%d", &l, &r, &val);
      update(0, N - 1, 1, l - 1, r - 1, val, 0);
    } else {
      scanf("%d%d", &l, &val);
      update(0, N - 1, 1, l - 1, l - 1, val, 1);
    }
  }
  for (int i = 0; i < an.size(); i++) cout << an[i] << "\n";
  return 0;
}